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K16288
CIVIL ENGINEERING
Jai B. KimRobert H. Kim
Jonathan R. Eberle
SIMPLIFIED LRFD BRIDGE DESIGN
SIMPLIFIED
LRFD BRID
GE D
ESIGN
Kim • Kim
• Eberle
With Eric J. Weaver and Dave M. Mante
Simplified LRFD Bridge Design is a study guide for solving bridge problems on the Civil and Structural PE exams. It is also suitable as a reference for practicing engineers and as a classroom text for civil engineering students. The book conforms to the fifth edition of AASHTO LRFD Bridge Design Specifications (2010).
Unlike most engineering books, Simplified LRFD Bridge Design uses an alternative approach to learning––the inductive method. The book introduces topics by presenting specific design examples, literally teaching backward––the theory is presented once specific design examples are comprehended.
Another unique quality of the book is that whenever new topics and materials appear in design examples, AASHTO LRFD Bridge Design Specifications reference numbers are cited, so that students will know where to find those new topics and materials.
For example,
New Topics or Material AASHTO Reference Number Cited
Design Live Load HL-93 A Art. 3.6.1.2
Design Examples and Practice ProblemsIn addition to the first section on an overview of the LRFD Method of Bridge Design, there are eight design examples and three practice problems utilizing a step-by-step process to help students learn easily in the shortest time.
About the EditorsJai B. Kim, PhD, PE, is a professor emeritus of civil and environmental engineering at Bucknell University, and was department chairman for 26 years. Recently he was a structural engineer at FHWA. He was also actively involved in the NCEES structural PE Committee and Transportation Research Board Committee of Bridges and Structures. He holds a BSCE and MSCE from Oregon State University and a PhD from University of Maryland. Robert H. Kim, MSCE, PE, is chief design engineer for BKLB Structural Consultants, Inc. He has extensive experience in bridge engineering. He holds a BS from Carnegie Mellon University and a MSCE from The Pennsylvania State University. Jonathan R. Eberle, BSCE and EIT, is engaged in research with a focus on the seismic design of structures at Virginia Polytechnic Institute. He holds a BSCE from Bucknell University.
SIMPLIFIED LRFDBRIDGE DESIGN
Boca Raton London New York
CRC Press is an imprint of theTaylor & Francis Group, an informa business
Jai B. KimRobert H. Kim
Jonathan R. Eberle
SIMPLIFIED LRFDBRIDGE DESIGN
With Eric J. Weaver and Dave M. Mante
CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742
© 2013 by Taylor & Francis Group, LLCCRC Press is an imprint of Taylor & Francis Group, an Informa business
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v© 2008 Taylor & Francis Group, LLC
Contents
List of Figures ....................................................................................................... viiList of Tables ........................................................................................................ xiiiPreface .....................................................................................................................xvAcknowledgments ............................................................................................. xviiIntroduction ......................................................................................................... xixEditors ................................................................................................................. xxiiiContributors ........................................................................................................ xxvNomenclature ...................................................................................................xxvii
1 LRFD Method of Bridge Design .................................................................1Limit States .......................................................................................................1Load Combinations and Load Factors ..........................................................3Strength Limit States for Superstructure Design ........................................6Resistance Factors, Φ, for Strength Limits ....................................................6Design Live Load HL-93 .................................................................................7Fatigue Live Load ............................................................................................7Number of Design Lanes, NL .........................................................................9Multiple Presence Factor of Live Load, m ....................................................9Dynamic Load Allowance, IM ..................................................................... 10Live Load Distribution Factors .................................................................... 10Load Combinations for the Strength I Limit State .................................... 11Simple Beam Live Load Moments and Shears Carrying Moving
Concentrated Loads per Lane ................................................................. 13Live Load Moments and Shears for Beams (Girders) ............................... 13
2 Design Examples........................................................................................... 17Design Example 1: Reinforced Concrete T-Beam Bridge ......................... 17Design Example 2: Load Rating of Reinforced Concrete T-Beam
by the Load and Resistance Factor Rating (LRFR) Method ................ 61Design Example 3: Composite Steel–Concrete Bridge ............................. 74Design Example 4: Longitudinal Steel Girder ......................................... 118Design Example 5: Reinforced Concrete Slabs ........................................ 144Design Example 6: Prestressed Interior Concrete Girder ...................... 164Design Example 7: Flexural and Transverse Reinforcement
for 50 ft Reinforced Concrete Girder .................................................... 183Design Example 8: Determination of Load Effects Due to Wind
Loads, Braking Force, Temperature Changes, and Earthquake Loads Acting on an Abutment .............................................................. 201
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3 Practice Problems ........................................................................................ 215Practice Problem 1: Noncomposite 60 ft Steel Beam Bridge for
Limit States Strength I, Fatigue II, and Service .................................. 215Practice Problem 2: 161 ft Steel I-Beam Bridge with Concrete Slab ...... 244Practice Problem 3: Interior Prestressed Concrete I-Beam .................... 263
Appendix A: Distribution of Live Loads per Lane for Moment in Interior Beams (AASHTO Table 4.6.2.2.2b-1) ................ 287
Appendix B: Distribution of Live Loads per Lane for Moment in Exterior Longitudinal Beams (AASHTO Table 4.6.2.2.2d-1) ................................................. 291
Appendix C: Distribution of Live Load per Lane for Shear in Interior Beams (AASHTO Table 4.6.2.2.3a-1) ................. 295
Appendix D: Distribution of Live Load per Lane for Shear in Exterior Beams (AASHTO Table 4.6.2.2.3b-1)................ 299
Appendix E: U.S. Customary Units and Their SI Equivalents ................303
References ...........................................................................................................305Primary References .....................................................................................305Supplementary References .........................................................................305
vii© 2008 Taylor & Francis Group, LLC
List of Figures
Figure 1.1 Design truck (HS-20), design tandem load (a pair of 25 kip axles 4 ft apart), and design lane load (0.64 kips/ft longitudinally distributed) ...............................................................................................................8
Figure 1.2 Fatigue live loading ..........................................................................9
Figure 1.3 Shear and moment diagrams for controlling design truck (HS-20) live load position ..................................................................................... 14
Figure 1.4 Shear and moment diagrams for the design truck (HS-20) center axle at midspan .......................................................................................... 15
Figure 2.1 T-beam design example ................................................................. 18
Figure 2.2 Interior T-beam section .................................................................. 20
Figure 2.3a Influence lines for moment at midspan ..................................... 21
Figure 2.3b Influence lines for shear at support ........................................... 21
Figure 2.4 Design truck (HS-20) position for moment at midspan ............22
Figure 2.5 Design tandem load position for moment at midspan..............22
Figure 2.6 Design truck (HS-20) position for shear at support ...................22
Figure 2.7 Design tandem load position for shear at support ....................22
Figure 2.8 Lever rule for determination of distribution factor for moment in exterior beam, one lane loaded .......................................................25
Figure 2.9 Moment distribution for deck slab and wearing surface loads .. 32
Figure 2.10 Moment distribution for curb and parapet loads for exterior girder ..................................................................................................33
Figure 2.11 T-beam section and reinforcement in T-beam stem .................36
Figure 2.12 Critical shear section at support ................................................. 41
Figure 2.13 Lane load position for maximum shear at critical shear section .....................................................................................................................43
Figure 2.14 Fatigue truck loading and maximum moment at 32 kips position per lane due to fatigue loading ............................................................ 51
Figure 2.15 Cracked section determination of T-beam ................................54
Figure 2.16 T-beam bridge cross section ........................................................ 62
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Figure 2.17 T-beam section ............................................................................... 62
Figure 2.18 Interior T-beam section for determination of flexural resistance ................................................................................................................ 67
Figure 2.19 Critical section for shear at support ........................................... 69
Figure 2.20 Live load shear at critical shear section due to lane load ........ 70
Figure 2.21 Composite steel–concrete bridge example ................................75
Figure 2.22 Steel section ...................................................................................75
Figure 2.23 Composite steel section ................................................................ 76
Figure 2.24 Composite section for stiffness parameter, Kg.......................... 79
Figure 2.25 Lever rule for determination of distribution factor for moment in exterior beam, one lane loaded ................................................. 81
Figure 2.26 Load position for moment at midspan for design truck load (HS-20) ............................................................................................................83
Figure 2.27 Load position for moment at midspan for design tandem load ...........................................................................................................84
Figure 2.28 Load position for moment at midspan for design lane load ..84
Figure 2.29 Load position for shear at support for design truck load (HS-20) ....................................................................................................................85
Figure 2.30 Load position for shear at support design tandem load .........85
Figure 2.31 Load position for shear at support for design lane load .........86
Figure 2.32 Composite steel–concrete section for shear and moment capacity calculation ...............................................................................................88
Figure 2.33 Section and cross section of interior girder for plastic moment capacity ................................................................................................... 91
Figure 2.34 Composite cross section for exterior beam ............................... 92
Figure 2.35 Section and cross section of exterior girder for plastic moment capacity ................................................................................................... 95
Figure 2.36 Interior girder section prior to transformed area .................. 100
Figure 2.37 Interior girder section after transformed area ....................... 101
Figure 2.38 Dimensions for transformed interior beam section .............. 101
Figure 2.39 Exterior girder section prior to transformed area.................. 106
Figure 2.40 Exterior girder section after transformed area ....................... 106
Figure 2.41 Dimensions for transformed exterior beam section .............. 107
ixList of Figures
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Figure 2.42 Single lane fatigue load placement with one design truck load for maximum moment at midspan .......................................................... 112
Figure 2.43 Single lane fatigue load placement with one design truck load for maximum shear at support ................................................................. 116
Figure 2.44 Steel girder bridge, 40 ft span ................................................... 118
Figure 2.45a Influence line for maximum moment at midspan ............... 121
Figure 2.45b Controlling load position for moment at midspan for design truck load (HS-20) ............................................................................ 121
Figure 2.45c Controlling load position for moment at midspan for design tandem load ...................................................................................... 121
Figure 2.45d Controlling load position for moment at midspan for design lane load............................................................................................. 121
Figure 2.45e Single lane fatigue load placement with one design truck load for maximum moment at midspan ............................................... 122
Figure 2.46 Noncomposite steel section at midspan .................................. 123
Figure 2.47 Lever rule for determination of distribution factor for moment in exterior beam, one lane loaded ............................................... 125
Figure 2.48a Maximum live load shears; influence line for maximum shear at support ................................................................................................... 127
Figure 2.48b Controlling load position for shear at support for design truck load (HS-20) ............................................................................ 127
Figure 2.48c Controlling load position for shear at support for design tandem load ...................................................................................... 127
Figure 2.48d Controlling load position for shear at support for design lane load............................................................................................. 127
Figure 2.48e Single lane fatigue load placement with one design truck load for maximum shear at support ...................................................... 128
Figure 2.49 Position of design truck loading (HS-20) for deflection at midspan ............................................................................................................ 137
Figure 2.50 Concrete deck slab design example ......................................... 145
Figure 2.51 Locations in slab strips for maximum reactions and moments due to dead loads ............................................................................... 146
Figure 2.52 Moments and reactions for deck slab dead load excluding deck cantilever .................................................................................. 147
Figure 2.53 Moments and reaction for deck slab dead load in deck cantilever .............................................................................................................. 147
x List of Figures
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Figure 2.54 Moments and reaction for curb and parapet loads ............... 148
Figure 2.55 Moments and reaction for wearing surface loads .................. 148
Figure 2.56 Live load placement for maximum negative moment ........... 149
Figure 2.57 Live load placement for maximum negative moment, one lane loaded .................................................................................................... 150
Figure 2.58 Live load placement for maximum positive moment in first interior span, one lane loaded .............................................................. 151
Figure 2.59 Live load placement for maximum positive moment, double lane loaded .............................................................................................. 152
Figure 2.60 Live load placement for maximum negative moment in first interior span, one lane loaded .............................................................. 153
Figure 2.61 Live load placement for maximum reaction at first support .. 154
Figure 2.62 Deck slab section for reinforcement placement...................... 158
Figure 2.63 Prestressed concrete interior girder design example ............ 165
Figure 2.64 Cross section of girder with composite deck .......................... 166
Figure 2.65 Area transformed section of girder section ............................ 168
Figure 2.66 Bending moments at midspan due to HL-93 loading ........... 171
Figure 2.67 Girder I-beam section ................................................................. 176
Figure 2.68 Concrete stresses at midspan at release of prestress for girder I-beam ................................................................................................. 179
Figure 2.69 Final concrete stresses at midspan after losses ...................... 183
Figure 2.70 Reinforced concrete girder design example ........................... 184
Figure 2.71 Girder section with area transformed deck slab .................... 186
Figure 2.72 Lever rule for distribution factor for exterior girder moment with one lane loaded ........................................................................... 190
Figure 2.73 Design truck (HS-20) load position for the maximum moment at midspan ............................................................................................ 192
Figure 2.74 Design lane load moment at midspan ..................................... 193
Figure 2.75 Design truck load (HS-20) load position for the maximum shear at support ............................................................................... 193
Figure 2.76 Design lane load position for the maximum shear at support ............................................................................................................. 194
Figure 2.77 Reinforcement details ................................................................ 196
xiList of Figures
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Figure 2.78 Review of shear reinforcement ................................................. 199
Figure 2.79 Abutment structure 16 ft in height and 29.5 ft in width ....... 202
Figure 2.80 Two-lane bridge supported by seven W30 × 108 steel beams at 29.5 ft wide abutment ......................................................................... 202
Figure 2.81 Steel beams at abutment and away from abutment .............. 203
Figure 2.82 Wind loads on abutment transmitted from superstructure ... 205
Figure 2.83 Wind loads on abutments transmitted from vehicle live load ................................................................................................................ 206
Figure 2.84 Forces on abutment from braking ............................................ 208
Figure 2.85 Summary of forces on abutment due to wind loads, braking forces, temperature changes, and earthquake loads ....................... 214
Figure 3.1 Cross section of noncomposite steel beam bridge .................... 216
Figure 3.2 W40 × 249 properties .................................................................... 216
Figure 3.3 Dead loads for interior girder ..................................................... 217
Figure 3.4 Dead loads for exterior girder ..................................................... 220
Figure 3.5 Section for longitudinal stiffness parameter, Kg ......................223
Figure 3.6 Lever rule for the distribution factor for moments for exterior girder ................................................................................................225
Figure 3.7 Design truck (HS-20) load moment at midspan .......................227
Figure 3.8 Design tandem load moment at midspan .................................227
Figure 3.9 Design lane load moment ............................................................228
Figure 3.10 Design truck (HS-20) shear at support .....................................228
Figure 3.11 Design tandem load shear at support ......................................229
Figure 3.12 Design lane load shear ...............................................................229
Figure 3.13 Center of gravity of design truck loading (HS-20) .................234
Figure 3.14 Fatigue load position for maximum moment .........................234
Figure 3.15 Fatigue load position for maximum shear ..............................235
Figure 3.16 Design truck loading for maximum deflection at midspan ... 240
Figure 3.17 Design lane loading for maximum deflection at midspan .... 242
Figure 3.18 Steel I-beam with concrete slab ................................................. 245
Figure 3.19 I-beam properties ........................................................................ 246
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Figure 3.20 Cross-section properties for shears and moments due to dead loads ........................................................................................................ 246
Figure 3.21 Design truck load (HS-20) position for maximum shear ......250
Figure 3.22 Design tandem load position for maximum shear ................250
Figure 3.23 Design lane load position for maximum shear ...................... 251
Figure 3.24 Design truck load (HS-20) position for maximum moment ... 251
Figure 3.25 Design tandem load position for maximum moment ........... 252
Figure 3.26 Design lane load position for maximum moment ................. 252
Figure 3.27 Lever rule for the distribution factor for moments for exterior girder ................................................................................................255
Figure 3.28 Fatigue load position for maximum moment at midspan .... 260
Figure 3.29 Prestressed concrete I-beam ......................................................264
Figure 3.30 Deck and I-beam .........................................................................264
Figure 3.31 Curb and parapet ........................................................................ 265
Figure 3.32 Composite section ....................................................................... 267
Figure 3.33 Influence line diagram for maximum moment at midspan ... 272
Figure 3.34 Design truck (HS-20) position for moment at midspan ........ 272
Figure 3.35 Design tandem load position for moment at midspan .......... 273
Figure 3.36 Design lane load for moment at midspan ............................... 273
Figure 3.37 Influence line diagram for maximum shear at support ........ 274
Figure 3.38 Design truck position for shear at support ............................. 274
Figure 3.39 Design tandem load position for shear at support ................ 274
Figure 3.40 Design lane load for shear at support ...................................... 274
Figure 3.41 Prestressed I-beam with 30, ½ in strands ................................ 279
xiii© 2008 Taylor & Francis Group, LLC
List of Tables
Table 1.1 Load Combinations and Load Factors .............................................4
Table 1.2 Load Factors for Permanent Loads, γp ..............................................5
Table 1.3 Multiple Presence Factors, m .............................................................9
Table 1.4 Dynamic Load Allowance, IM ........................................................ 10
Table 2.1 Distributed Live Load and Dead Load Effects for Interior Beam for Reinforced Concrete T-Beam Bridge ................................................. 31
Table 2.2 Unfactored Beam Moments and Shears Due to Dead Loads and Live Loads for Reinforced Concrete T-Beam Bridge ................................34
Table 2.3 Dead Loads and Distributed Live Loads Effects Summary for Interior T-Beam ................................................................................................ 71
Table 2.4 Load Factors for Load Rating for Reinforced Concrete Bridge ...72
Table 2.5 Rating Factor (RF) ............................................................................. 73
Table 2.6 Complete Live Load Effect Summary ............................................86
Table 2.7 Summary of Plastic Moment Capacity and Shear Force ............. 96
Table 2.8 Load Modifier Factors .................................................................... 120
Table 2.9 Summary of Distribution Factors: ................................................ 129
Table 2.10 Summary of Fatigue Limit State Distribution Factors ............ 129
Table 2.11 Summary of Loads, Shears, and Moments in Interior Beams .. 132
Table 2.12 Summary of Loads, Shears, and Moments in Exterior Beams .................................................................................................................... 133
Table 2.13 Force Effects Summary Table ...................................................... 154
Table 2.14 Load Factors for Permanent Loads ............................................. 155
Table 2.15 Strength I Limit State Summary ................................................. 157
Table 2.16 Summary of Section Properties .................................................. 169
Table 2.17 Unfactored Moments per Girder ................................................ 173
Table 2.18 Summary of Distribution Factors ............................................... 191
Table 2.19 Summary of Forces ....................................................................... 213
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Table 3.1 Dead Load Summary of Unfactored Shears and Moments for Interior and Exterior Girders .......................................................................222
Table 3.2 Summary of Live Load Effects ......................................................230
Table 3.3 Summary of Dead Load Shears and Moments in Interior Beams .................................................................................................250
Table 3.4 Summary of Live Load Shears and Moments ............................253
Table 3.5 Summary of Live Load Distribution Factors ..............................258
Table 3.6 Summary of Unfactored Distributed Live Load Effects per Beam ............................................................................................................... 259
Table 3.7 Load Modifiers ................................................................................ 265
Table 3.8 Dynamic Load Allowance, IM ...................................................... 270
Table 3.9 Summary of Dead Load Moments and Shears ........................... 276
xv© 2008 Taylor & Francis Group, LLC
Preface
We have developed this first edition of LRFD Bridge Design to comply with the fifth edition of AASHTO LRFD Bridge Design Specification (2010) (referred to as AASHTO throughout this book), which should be at your side while using this guide. This book primarily serves engineers studying to take the National Council of Examiners for Engineering and Surveying (NCEES) structural PE examination and the structural depth section of the NCEES civil PE exam. It is also suitable as a classroom text for civil engineering seniors and graduate students, as well as a reference book for practicing engineers.
xvii© 2008 Taylor & Francis Group, LLC
Acknowledgments
We wish to acknowledge Bucknell University, located in Lewisburg, Pennsylvania, for providing a conducive environment for the writing of this book, and the students in our CENG 461/661 LRFD Bridge Design for their comments and suggestions. Thanks in particular to students Craig Stodart, EIT and Dale Statler, EIT. Also those former students who made significant contributions to this book are Eric J. Weaver, Dave M. Mante, and Jonathan R. Eberle.
Finally we dedicate this book to Yung J. Kim, mother of Robert and wife of Jai B. Kim, who provided the necessary support and motivation for this book.
Should you find an error here, we hope two things happen: first, that you will let us know about it; and second, that you will learn something from the error; we know we will! We would appreciate constructive comments, sug-gestions for improvement, and recommendations for expansion.
Good luck on the exam!
xix© 2008 Taylor & Francis Group, LLC
Introduction
The primary function of Simplified LRFD Bridge Design is to serve as a study reference for practicing engineers and students preparing to take the National Council of Examiners for Engineering and Surveying (NCEES) civil and structural exams. As such, this book guides you through the application of the fifth (2010) edition of the AASHTO LRFD Bridge Design Specifications, which you must have at your side as you work this book’s problems.
Be aware that although AASHTO is incorporated into many major build-ing codes and structural specifications, there may be codes and specifica-tions that differ from, and take priority over, the specifications in AASHTO. In practice you should check with the governing jurisdiction to confirm which codes and specifications must be followed. In addition to AASHTO, you may need to consult other references for more comprehensive explana-tions of bridge design theory.
This book’s first chapter, “LRFD Method of Bridge Design,” intro-duces you to the key steps of LRFD bridge design as they relate to the book’s eight design examples and three practice problems. The chapter also includes and describes the use of many key tables and figures from AASHTO. Because this book covers various AASHTO subjects, you may use it to brush up on a few specific subjects, or may study the book in its entirety. Do note, however, that the eight design examples are the most exhaustive in their applications of AASHTO subjects, and that the three practice problems that follow build on concepts and information that have been set out in those first eight examples. You can use this book most effectively by studying the design examples in order. Furthermore, the book’s explanations are meant to explain and clarify AASHTO; however, they assume that the reader can refer directly to AASHTO itself when necessary. Among the book’s examples are references to AASHTO tables (“A Tbl . . .”), sections (“A Sec . . .”), figures (“A Fig . . .”), and equations (“A Eq . . .”).
Throughout the book, example and practice problems illustrate How To Use the AASHTO LRFD Bridge Design Specifications, fifth edition (2010). Take your time with these and make sure you understand each example before moving ahead. Keep in mind, though, that in actual design situations there are often several correct solutions to the same problem.
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If You Are a Practicing Engineer, Engineering Student, or Instructor
Although this book is primarily intended to aid in exam preparation, it is also a valuable aid to engineers, and can serve as a classroom text for civil engineering seniors and graduate students. For anyone using this book, the design examples serve as a step-by-step, comprehensive guide to bridge design using AASHTO.
If You Are an Examinee
If you are preparing to take the NCEES civil, or structural PE exam, work all of the examples in this book to prepare yourself on the application of the principles presented. By solving the problems in this book you will have a better understanding of the elements of bridge design that could be part of the problems on the exams. By reviewing the solutions, you will learn effi-cient problem-solving methods that may benefit you in a timed exam.
About the Exams
In April 2011, the new 16-hour structural exam replaced the separate Structural I and II exams. The new exam is a breadth and depth exam offered in two components on successive days. The eight-hour Vertical Forces (Gravity/Other) and Incidental Lateral component is offered only on Friday and focuses on gravity loads and lateral earth pressures. The eight-hour Lateral Forces (Wind/Earthquake) component is offered only on Saturday and focuses on wind and earthquake loads.
Each component of the SE exam has a breadth (morning) and a depth (afternoon) module. Examinees must take the breadth module of each com-ponent and one of the two depth modules in each component.
Breadth modules (morning sessions): These modules contain questions covering a comprehensive range of structural engineering topics. All questions are multiple choice.
Depth modules (afternoon sessions): These modules focus more closely on a single area of practice in structural engineering. Examinees must choose either buildings or bridges. Examinees must work the same topic area on both components.
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The civil PE exam consists of two sessions, each lasting four hours and consisting of 40 multiple choice questions, but the questions in the morning and afternoon sessions are of about equal difficulty. The morning (breadth) session of the exam may contain general bridge design–related problems. The structural afternoon (depth) session of the exam may include more in-depth bridge design–related problems. The problems in each session typi-cally require an average of six minutes to work
Although the format of the design examples presented in this book dif-fers from those six-minute problems for the civil and structural PE exams, as you work the problems in this book in preparation for either the civil or structural exam, you will find all of the topics covered here also covered on the structural PE exam in some form or another. Using this book will help you gain a broader knowledge base and understanding of the many bridge design subjects covered on exams.
xxiii© 2008 Taylor & Francis Group, LLC
Editors
Jai B. Kim, PE, PhD, is a professor emeritus of civil and environmental engineering at Bucknell University. He was department chairman for 26 years. Also, currently he is a bridge consultant (since 1980) and president of BKLB Structural Consultants, Inc. Recently he was a structural engi-neer at the Federal Highway Administration (FHWA). He has been active in bridge research for over 40 years, and is currently a member of the Transportation Research Board Committee of Bridges and Structures. He also served on the Structural PE Exam Committee of the National Council of Examiners and Surveying (NCEES) for many years. He holds a BSCE and MSCE from Oregon State University and a PhD from the University of Maryland.
Robert H. Kim, PE, MSCE, is chief design engineer for BKLB Structural Consultants, Inc. He has extensive experience in the design, research, and construction of highway bridges. He has authored and presented several papers related to bridge engineering. Robert’s three books, Bridge Design for the Civil and Structural Professional Exams, Second Edition; Timber Design, Seventh Edition; and Civil Discipline Specific Review for the FE/EIT Exam are well read by both students and engineers. In 2013, he is working on a bridge rehabilitation design in Connecticut. He holds a BS from Carnegie Mellon University and a MSCE from The Pennsylvania State University.
Jonathan R. Eberle, BSCE, is engaged in research with focus on the seismic design and analysis of structures at Virginia Polytechnic Institute and State University as a graduate student. He holds a BSCE from Bucknell University.
xxv© 2008 Taylor & Francis Group, LLC
Contributors
David M. Mante, BSCE, performed a rigorous full-scale laboratory test-ing program focused on developing and testing of an innovative concrete bridge deck system. Presently, as a PhD student at Auburn University, he is a guest lecturer in undergraduate civil engineering courses and actively performs research related to prestressed concrete bridge girders. He holds a BSCE from Bucknell University.
Eric J. Weaver, PE, M.ASCE, M.ASME, graduated from Bucknell University with a BS in civil engineering and earned an MEng in struc-tural engineering from Lehigh University, where the primary focus of his research was fatigue and life-cycle analysis of steel truss bridges. Following graduate school, Eric worked for several years as a design engi-neer on NASA’s Space Shuttle program and is currently employed as a structural engineer for Westinghouse Electric Company.
xxvii© 2008 Taylor & Francis Group, LLC
Nomenclature
Symbol: Definition (Units)
A: bearing pad area (in2)A: area of stringer, beam, or girder (in2)a: depth of equivalent rectangular stress block (in)A1: factor for dead load used in computing the rating factorA2: factor for live load used in computing the rating factorAb: area of concrete reinforcing bar (in2)Ac: area of composite section (in2)ADT: average daily traffic (vehicles/day)ADTT: average daily truck trafficADTTSL: single-lane average daily truck trafficAg: gross area of cross-section (in2)Agc: area of transformed gross composite section (in2)Aps: area of prestressing steel (in2)As: area of nonprestressed reinforcement (in2)As: peak seismic ground acceleration coefficient modified by short-period
site factorAs,temp: area of temperature reinforcement in concrete slab (in2)Av: area of transverse reinforcement with distance s (in2)b: width of beam or width of the compression face of the member (in)bc: width of the compression flange (in)be: effective flange width for beams (in)bet: transformed effective deck width (in)bf: full width of the flange (in)bi: flange width of interior beam (in)bmin: minimum width of T-beam stem (in)BR: vehicular braking force (kips)BR: vertical braking force (kips/ft)BRhor: horizontal braking force at the top of the abutment (kips/ft)BRmax: maximum braking force (kips)BRtandem: braking force resulting from tandem, single traffic lane (kips)BRtandem+lane: braking force resulting from tandem and lane load, single traffic
lane (kips)BRtruck: braking force resulting from truck, single traffic lane (kips)
xxviii Nomenclature
© 2008 Taylor & Francis Group, LLC
BRtruck+lane: braking force resulting from truck and lane load, single traffic lane (kips)
BRvert: vertical braking force at the top of the abutment (kips/ft)bs: effective width of concrete deck (in)bs: width of beam (in)bs,ext: effective flange width for exterior beams (in)bs,int: effective flange width for interior beams (in)bt: width of the tension flange (in)bf: flange width of steel beam section (in)bv: width of web (in)BW: barrier weight (kips/ft)bw: web width (in)c: distance from the extreme compression fiber to the neutral axis (in)C: ratio of the shear buckling resistance to the shear specified minimum
yield strengthC: stiffness parameterC&P: curb and parapet cross-section area (ft2)c.g.: center of gravityCE: vehicular centrifugal forceCL: center lineCR: forces resulting from creepCrb: distance from top of concrete deck to bottom layer of longitudinal con-
crete deck reinforcement (in)Crt: distance from top of concrete deck to top layer of longitudinal concrete
deck reinforcement (in)CT: vehicular collision forceCV: vessel collision forceD: clear distance between flanges (in)D: dead load (lbf)D: depth of steel beam (in)D: width of distribution per lane (ft)d: depth of beam or stringer (in)db: nominal diameter of reinforcing bar, wire, or prestressing strand (in)dc: concrete cover measured from extreme tension fiber to the center of the
flexural reinforcement located closest thereto (in)dc: distance from the compression flange to the PNA (in)DC: dead load of structural components and nonstructural attachments (kips)DC1: noncomposite dead load (kips/ft)DC2: composite dead load (kips/ft)DCC&P: distributed load resulting from curb and parapet self-weight (kips/ft)DChaunch: noncomposite dead load resulting from haunch self-weight (kips/ft)Dcp: depth of girder web in compression at the plastic moment (in)DCslab: noncomposite dead load resulting from slab self-weight (kips/ft)DCstay-in-place forms: noncomposite dead load resulting from self-weight of
stay-in-place forms (kips/ft)
xxixNomenclature
© 2008 Taylor & Francis Group, LLC
DCT-beam: distributed load resulting from T-beam self-weight (kips/ft)DD: downdrag loadde: effective depth from extreme compression fiber to the centroid of the ten-
sile force in the tensile reinforcement (in)de: horizontal distance from the centerline of the exterior web of exterior
beam at the deck level to the interior edge of curb at barrier.DF: distribution factor for moment or shearDFdeflection: distribution factor for deflectionDFM: distribution factor for momentDFME
fat: load distribution for fatigue moments, exterior girderDFMext: load distribution for moments, exterior girdersDFMfat,ext: load distribution for fatigue moments, exterior girderDFMfat,int: load distribution for fatigue moments, interior girderDFMfatigue: load distribution for fatigue momentsDFMI
fat: load distribution for fatigue moments, interior girderDFMint: load distribution for moments, interior girdersDFMme: distribution factor for moment for multiple design lanes loaded for
exterior beamsDFMmi: distribution factor for moment for multiple design lanes loaded for
interior beamsDFMse: distribution factor for moment for a single design lane loaded for
exterior beamsDFMsi: distribution factor for moment for a single design lane loaded for
interior beamsDFV: distribution factor for shearDFVext: load distribution for shears, exterior girdersDFVfat,ext: load distribution for fatigue shears, exterior girderDFVfat,int: load distribution for fatigue shears, interior girderDFVint: load distribution for shears, interior girdersDFVme: distribution factor for shear for multiple design lanes loaded for
exterior beamsDFVmi: distribution factor for shear for multiple design lanes loaded for
interior beamsDFVse: distribution factor for shear for a single design lane loaded for exterior
beamsDFVsi: distribution factor for shear for a single design lane loaded for interior
beamsdgirder: depth of girder (in)do: transverse stiffener spacing (in)dp: distance from extreme compression fiber to the centroid of the prestress-
ing tendons (in)Dp: distance from the top of concrete deck to the neutral axis of the compos-
ite section (in)ds: distance from extreme compression fiber to the centroid of the nonpre-
stressed tensile reinforcement (in)
xxx Nomenclature
© 2008 Taylor & Francis Group, LLC
ds: thickness of concrete deck slab (in)Dt: depth of the composite section (in)dt: distance from the tension flange to the PNA (in)dv: effective shear depth (in)dw: distance from the web to the PNA (in)DW: superimposed dead load (wearing surfaces and utilities) (kips or kips/ft)DWFWS: future wearing surface dead load (kips/ft)e: correction factor for load distribution for exterior beamsE: modulus of elasticity of steel (ksi)EB: modulus of elasticity of beam material (kips/in2)Ebeam: modulus of elasticity of beam (ksi)Ec: modulus of elasticity of concrete (ksi)ec: strand eccentricity at midspan (in)Ecg: modulus of elasticity of concrete after 28 days (ksi)Eci: modulus of elasticity of concrete at transfer (ksi)Ecs: modulus of elasticity of concrete after losses (ksi)ED: modulus of elasticity of deck material (kips/in2)Edeck: modulus of elasticity of the deck (ksi)eg: distance between the centers of gravity of the beam and deck (in)EH: horizontal earth pressure loadEL: accumulated locked-in force effects resulting from the construction pro-
cess, including the secondary forces from posttensioningem: average eccentricity at midspan (in)Ep: modulus of elasticity of prestressing tendons (ksi)EQ: forces resulting from earthquake loading (kips)EQh: horizontal earthquake loading at the top of the abutment (kips/ft)ES: earth surcharge loadEs: modulus of elasticity of prestressing steel (kips/in2)Es: modulus of elasticity of steel (ksi)f: bending stress (kips/in2)f′c: compressive strength of concrete at 28 days (ksi)f′c, beam: beam concrete strength (kips/in2)f′c, deck: deck concrete strength (kips/in2)f′cg: compressive strength of concrete at 28 days for prestressed I-beams (ksi)f′cgp: the concrete stress at the center of gravity of prestressing tendons due
to prestressing force immediately after transfer and self-weight of member at section of maximum moment (ksi)
f′ci: compressive strength of concrete at time of prestressing transfer (ksi)f′cs: compressive strength of concrete at 28 days for roadway slab (ksi)f′s: stress in compression reinforcement (ksi)fbt: amount of stress in a single strand at 75% of ultimate stress (kips/in2)fbu: required flange stress without the flange lateral bendingfc: compressive stress in concrete at service load (ksi)fcgp: concrete stress at the center of gravity of prestressing tendons that
results from the prestressing force at either transfer or jacking and the self-weight of the member at sections of maximum moment (ksi)
xxxiNomenclature
© 2008 Taylor & Francis Group, LLC
fci: temporary compressive stress before losses due to creep and shrinkage (ksi)
fcpe: compressive stress in concrete due to effective prestress forces only (after allowance for all prestress losses) at extreme fiber of section where tensile stress is caused by externally applied loads (ksi)
fcs: compressive strength of concrete after losses (ksi)fDC: steel top flange stresses due to permanent dead loads (kips/in2)fDW: steel top flange stresses due to superimposed dead load (kips/in2)ff: flange stress due to the Service II loads calculated without consideration
of flange lateral bending (ksi)ff: allowable fatigue stress range (ksi)fgb: tensile stress at bottom fiber of section (kips)fl: flange lateral bending stress due to the Service II loads (ksi)fLL+IM: steel top flange stresses due to live load including dynamic load
allowance (kips/in2)fmin: minimum live load stress resulting from the fatigue load combined with
the permanent loads; positive if in tension (kips/in2)fpbt: stress in prestressing steel immediately prior to transfer (ksi)fpc: compressive stress in concrete (after allowance for all prestress losses) at
centroid of cross-section resisting externally applied loads (ksi)*
fpe: compressive stress in concrete due to effective prestress forces only (after allowance for all prestress losses) at extreme fiber of section where tensile stress is caused by externally applied loads (ksi)
fpga: seismic site factorfps: average stress in prestressing steel at the time for which the nominal
resistance of member is required (ksi)fpt: stress in prestressing steel immediately after transfer (ksi)fpu: specified tensile strength of prestressing steel (ksi)fpul: stress in the strand at the strength limit state (ksi)fpy: yield strength of prestressing steel (ksi)fr: modulus of rupture of concrete (psi)fs: stress in the mild tension reinforcement at the nominal flexural resistance
(ksi)fs: stress in the reinforcement (ksi)fs: stress in the reinforcement due to the factored fatigue live load (kips/in2)fse: effective steel prestress after losses (ksi)fsi: allowable stress in prestressing steel (ksi)fss: tensile stress in mild steel reinforcement at the service limit state (ksi)ft: excess tension in the bottom fiber due to applied loads (kips)ft: tensile stress at the bottom fiber of the T-beam (kips/in2)fti: temporary tensile stress in prestressed concrete before losses (ksi)fts: tensile strength of concrete after losses (psi)FWS: future wearing surface (in)
* In a composite member, fpc is resultant compressive stress at centroid of composite section.
xxxii Nomenclature
© 2008 Taylor & Francis Group, LLC
fy: specified minimum yield strength of reinforcing bars (ksi)Fy: specified minimum yield strength of steel (ksi)Fyc: specified minimum yield strength of the compression flange (kips/in2)Fyf: specified minimum yield strength of a flange (ksi)Fyt: specified minimum yield strength of the tension flange (kips/in2)Fyw: specified minimum yield strength of a web (ksi)g: centroid of prestressing strand pattern (in)g: distribution factorG: shear modulus of bearing pad elastomers (ksi)ginterior = DFVmi: distribution factor designation for interior girdersgM
ME: distribution factor for moment with multiple lanes loaded, exterior girder
gMMI: distribution factor for moment with multiple lanes loaded, interior
girdergM
SE: distribution factor for moment with single lane loaded, exterior girdergM
SI: distribution factor for moment with single lane loaded, interior girdergV
ME: distribution factor for shear with multiple lanes loaded, exterior girdergV
MI: distribution factor for shear with multiple lanes loaded, interior girdergV
SE: distribution factor for shear with single lane loaded, exterior girdergV
SI: distribution factor for shear with single lane loaded, interior girderH: average annual ambient relative humidity (%)h: depth of deck (in)h: overall depth or thickness of a member (in)Hcontr: load due to contraction (kips)hmin: minimum depth of beam including deck thickness (in)hparapet: height of parapet (in)Hrise: load due to expansion (kips)Htemp fall: horizontal force at the top of the abutment due to temperature fall
(kips/ft)Htemp,fall: horizontal load due to temperature fall (kips/ft)Hu: ultimate load due to temperature (kips)I: moment of inertia (in4)I: live load impact factorIc: composite section moment of inertia (in4)Ig: moment of inertia of gross concrete section about centroidal axis, neglect-
ing reinforcement (in4)IM: dynamic load allowanceIp: polar moment of inertia (in4)Ix: moment of inertia with respect to the x-axis (in4)Iy: moment of inertia with respect to the y-axis (in4)Iyc: moment of inertia of the compression flange of the steel section about the
vertical axis in the plane of the web (in4)Iyt: moment of inertia of the tension flange of the steel section about the verti-
cal axis in the plane of the web (in4)k: shear-buckling coefficient for webs
xxxiiiNomenclature
© 2008 Taylor & Francis Group, LLC
Kg: longitudinal stiffness parameter (in4)L: span length of beam (ft)LL: vehicular live load, TL + LNLN: design lane loadLS: live load surchargeM: bending moment about the major axis of the cross-section (in-kips)M: moment designationm: multiple presence factorMall, inv: allowable bending moment for inventory rating (ft-kips)Mall, opr: allowable bending moment for operating rating (ft-kips)MC: moment at midspanMcr: cracking moment (in-kips)MD: moment due to slab dead loadMDC: moment due to superstructure dead load (ft-kips)MDC, tot: moment for the total component dead load (kips)MDC1: unfactored moment resulting from noncomposite dead loads (ft-kips)MDC2: unfactored moment resulting from composite dead loads (ft-kips)MDW: moment due to superimposed dead load (ft-kips)ME
U, fat: factored fatigue design live load moment, exterior beam (ft-kips)ME
fat+IM: unfactored distributed fatigue live load moment with impact, exte-rior beam (ft-kips)
Mf: moment per lane due to fatigue load (in-kips)MF,fatigue: factored moment per beam due to FatigueI load (in-kips)Mfat,ext: unfactored distributed moment resulting from fatigue loading, exte-
rior girder (ft-kips)Mfat,int: unfactored distributed moment resulting from fatigue loading, inte-
rior girder (ft-kips)Mfat,LL: fatigue moment due to live load (ft-kips)Mfatigue: unfactored moment per beam due to fatigue load (in-kips)Mg: midspan moment due to beam weight (in-kips)mgSI,M: distribution of live load moment per lane with one design lane loaded
for interior beamsMI: multiple lane, interior designationmi, MI: two or more design lanes loaded, interior girderMI
fat+IM: unfactored distributed fatigue live load moment with impact, inte-rior beam (ft-kips)
MIU, fat: factored fatigue design live load moment, interior beam (ft-kips)
MLL+IM: total live load moment per lane including impact factor (ft-kips)Mln: lane load moment per lane (in-kips)MLN: unfactored live load moment per beam due to lane load (in-kips)Mmax: maximum dead load moment (ft-kips)Mn: nominal flexural resistance (in-kips)Mp: plastic moment capacity of steel girder (ft-kips)Mr: factored flexural resistance of a section in bending, ΦMn (in-kips)Ms: moment due to superimposed dead loads (ft-kips)
xxxiv Nomenclature
© 2008 Taylor & Francis Group, LLC
Mservice: total bending moment resulting from service loadsMtandem: tandem load moment per lane (ft-kips)MTL: unfactored live load moment per beam due to truck load (in-kips)Mtr: HS-20 truck load moment per lane (in-kips)Mu: factored design moment at section ≤ ΦMn (in-kips)n: modular ratio = Es/Ec or Ep/Ec
N: number of stress cycles over fatigue design lifen: number of stress cycles per truck passageNb: number of beams, stringers, or girdersNc: number of cells in a concrete box girderNg: number of girdersNL: number of design lanesp: fraction of truck traffic in a single laneP: total nominal shear force in the concrete deck for the design of the shear
connectors at the strength limit state (kips)PB: base wind pressure corresponding to a wind speed of 100 mphPB: base wind pressure specified in AASHTO (kips/ft2)Pc: plastic force in the compression flange (kips)PC&P: load for the curb and parapet for exterior girders (kips/ft)PD: design wind pressure (kips/ft2)Pe: effective prestress after losses (kips)PGA: peak seismic ground acceleration coefficient on rock (Site Class B)Pi: initial prestress force (kips)PL: pedestrian live loadPNA: plastic neutral axisPpe: prestress force per strand after all losses (kips)Ppi: prestress force per strand before transfer (kips)Ppt: prestress force per strand immediately after transfer (kips)Prb: plastic force in the bottom layer of longitudinal deck reinforcement (kips)Prt: plastic force in the top layer of longitudinal deck reinforcement (kips)Ps: plastic force in the slab (kips)Pt: plastic force in the tension flange (kips)Pw: plastic force in the web (kips)Q: total factored load (kips)Qi: force effectQi: force effect from various loadsR: reaction at support (kips)RF: rating factor for the live load carrying capacityRF: rating factor for the live load carrying capacityRh: hybrid factorRn: nominal resistanceRr: factored resistance (ΦRn)RT: load rating for the HS-20 load at the inventory level (tons)S: section modulus of section (in3)s: spacing of bars or stirrups (in)
xxxvNomenclature
© 2008 Taylor & Francis Group, LLC
S: spacing of beams or webs (ft)S: spacing of supporting elements (ft)Sb, St: noncomposite section moduli (in3)Sbc, Stc: section moduli of composite beam section at the bottom and top
extreme fibers, respectively (in3)Sbottom: section modulus of the bottom steel flange (in3)Sc: section modulus for the extreme fiber of the composite section where ten-
sile stress is caused by externally applied loads (in3)Sc, Sbc: composite section moduli where the tensile stress is caused by exter-
nally applied loads (in3)Se: effective span length (ft)SE: loads resulting from settlementSE: single lane, exterior designationse, SE: single design lane loaded, exterior girderSg: section modulus for gross sectionSH: loads resulting from shrinkageSI: single lane, interior designationsi, SI: single design lane loaded, interior girdersmax: maximum spacing of flexural reinforcement (in)Snc: section modulus for the extreme fiber of the monolithic or noncomposite
section where tensile stress is caused by externally applied loads (in3)Snc,bottom: section modulus for extreme bottom fiber of the monolithic or
noncomposite section where tensile stress is caused by externally applied loads (in3)
Snc,top: section modulus for extreme top fiber of the monolithic or noncom-posite section where compressive stress is caused by externally applied loads (in3)
Sncb: section modulus for extreme bottom fiber of the monolithic or noncom-posite section where tensile stress is caused by externally applied loads (in3)
Snct: section modulus for extreme top fiber of the monolithic or noncompos-ite section where compressive stress is caused by externally applied loads (in3)
Stop: section modulus for the top flange (in3)Sx: section modulus with respect to the x-axis (in3)Sxt: elastic section modulus about the major axis of the section to the tension
flange (in3)Sxx: section modulus with respect to the y-axis (in3)Sy: section modulus with respect to the y-axis (in3)t: slab thickness (in)tbearing: thickness of bearing (in)tc: thickness of a compression flange (in)td: deck thickness (in)tdeck: thickness of deck (in)tf: flange thickness (in)
xxxvi Nomenclature
© 2008 Taylor & Francis Group, LLC
tg: depth of steel girder or corrugated steel plank including integral concrete overlay or structural concrete component, less a provision for grind-ing, grooving, or wear (in)
TG: loads resulting from temperature gradientTL: design truck load, or design tandem loadtmin: minimum depth of concrete slab to control deflection (in)to: depth of structural overlay (in)ts: thickness of concrete slab (in)tt: thickness of the tension flange (in)TU: loads resulting from uniform temperaturetw: web thickness (in)U: factored force effectV: shear designationV: shear force (kips)VB: base design wind velocity (mph)Vc: shear resistance provided by the concrete (kips)Vcr: shear-buckling resistance (kips)VDC: shear due to superstructure dead load (kips)VDC, tot: shear for the total component dead load (kips)VDC1: unfactored shear resulting from noncomposite dead loads (kips)VDC2: unfactored shear resulting from composite dead loads (kips)VDL: unfactored shear force caused by DL (kips)VDW: shear due to superimposed dead load (kips)VDZ: design wind velocity (mph)Vfat: shear force resulting from fatigue load (kips)Vfatigue: fatigue load shear per lane (kips)Vfatigue+IM: fatigue load shear per girder (kips)VLL+IM: total live load shear per lane including impact factor (ft-kips)Vln: lane load shear per lane (kips)VLN: unfactored live load shear per beam due to lane load (kips)Vmax: maximum dead load shear (kips)Vn: nominal shear resistance (kips)Vp: plastic shear resistance of the web (kips)Vp: shear yielding of the web (kips)Vpermanent: shear due to unfactored permanent load (kips)Vs: shear resistance provided by shear reinforcement (kips)Vtandem: tandem load shear per lane (kips)VTL: unfactored live load shear per beam due to truck load (kips)Vtr: truck load shear per lane (kips)Vu: factored shear force at section (kips)vu: average factored shear stress on concrete (ksi)Vu,ext: factored shear force at section in external girder (kips)Vu,int: factored shear force at section in internal girder (kips)Vu,total: total factored shear force at section (kips)w: distributed load (kips/ft2)
xxxviiNomenclature
© 2008 Taylor & Francis Group, LLC
W: weight in tons of truck used in computing live load effectw: width of clear roadway (ft)WA: water load and stream pressurewc: self-weight of concrete (kips/ft3)wC&P: distributed load resulting from self-weight of curb and parapet (kips/ft)wDC: distributed load of weight of supported structure (kips/ft2)wDW: distributed load of superimposed dead load (kips/ft2)wFWS: future wearing surface load (kips/ft2)WL: loads resulting from wind forces on live loadWL: wind pressure on vehicles, live loadWLh: horizontal loading due to wind pressure on vehiclesWLh: horizontal wind loading at the top of the abutment (kips/ft)WLv: vertical wind loading at the top of the abutment (kips/ft)ws: superimposed dead loads, parapet/curb load plus the future wearing
surface load (kips/ft2)WS: wind load pressure on superstructureWS: wind pressures on superstructures (kips)WSh: horizontal load on top of abutment due to wind pressure on
superstructureWSh: horizontal wind loading at the top of the abutment (kips/ft)wslab: distributed load of concrete slab (kips/ft2)wslab,ext: deck slab distributed load acting on exterior girder (kips/ft)wslab,int: deck slab distributed load acting on interior girder (kips/ft)WSsub: horizontal wind load applied directly to the substructureWStotal: total longitudinal wind loading (kips)WSv: vertical load on top of abutment due to wind pressure on superstructureWSv: vertical wind loading along the abutment (kips/ft)X: distance from load to point of support (ft)x: distance from beam to critical placement of wheel load (ft)x: distance of interest along beam span (ft)y′t, y′b, yt, and yb: for composite beam cross-section (in)yb: distance from the bottom fiber to the centroid of the section (in)ybs: distance from the center of gravity of the bottom strands to the bottom
fiber (in)yt: distance from the neutral axis to the extreme tension fiber (in)yt, yb: distance from centroidal axis of beam gross section (neglecting rein-
forcement) to top and bottom fibers, respectively (in)Zreq’d: required plastic section modulus (in3)α: angle of inclination of stirrups to longitudinal axisα: angle of inclination of transverse reinforcement to longitudinal axis (deg)β: factor indicating ability of diagonally cracked concrete to transmit tensionβ1: factor for concrete strengthβ1: ratio of the depth of the equivalent uniformly stressed compression zone
assumed in the strength limit state to the depth of the actual com-pression zone
xxxviii Nomenclature
© 2008 Taylor & Francis Group, LLC
βs: ratio of the flexural strain at the extreme tension face to the strain at the centroid of the reinforcement layer nearest the tension face
γ: load factorγe: exposure factorγh: correction factor for relative humidity of the ambient airγi: load factor; a statistically based multiplier applied to force effects includ-
ing distribution factors and load combination factorsγp: load factors for permanent loadsγst: correction factor for specified concrete strength at the time of the pre-
stress transfer to the concreteδ: beam deflection (in)Δ25% truck: 25% of deflection resulting from truck loading (in)Δ25% truck+ lane: 25% of deflection resulting from truck loading plus deflection
resulting from lane loading (in)Δcontr: contraction resulting from thermal movement (in)Δcontr: contractor thermal movementΔexp: expansion resulting from thermal movement (in)Δexp: expansion thermal movementΔfext: maximum stress due to fatigue loads for exterior girders (kips/in2)Δfint: maximum stress due to fatigue loads for interior girders (kips/in2)(Δf): load-induced stress range due to fatigue load (ksi)(ΔF)n: nominal fatigue resistance (ksi)ΔfpES: sum of all losses or gains due to elastic shortening or extension at the
time of application of prestress and/or external loads (ksi)ΔfpLT: losses due to long-term shrinkage and creep of concrete, and relax-
ation of the steel (ksi)ΔfpR: estimate of relaxation loss taken as 2.4 kips/in2 for low relaxation strand,
10.0 kips/in2 for stress-relieved strand, and in accordance with manu-facturer’s recommendation for other types of strand (kips/in2)
ΔfpT: total loss (ksi)(ΔF)TH: constant amplitude (ksi)Δtruck: deflection resulting from truck loading (in)δLL: deflection due to live load per lane (in)δLL+IM: deflection due to live load per girder including impact factor (in)δln: deflection due to lane load (in)δmax: maximum deflection for vehicular load (in)εx: tensile strain in the transverse reinforcementη: load modifierηD: ductility factor (strength only)ηi: load modifier relating to ductility redundancy, and operational impor-
tance = 1.0 (for conventional designs)ηI: operational importance factor (strength and extreme only) = 1.0 for (for
conventional bridges)ηR: redundancy factorθ: angle of inclination of diagonal compressive stresses (degrees)
xxxixNomenclature
© 2008 Taylor & Francis Group, LLC
Φ: resistance factorΦc: condition factorΦf: resistance factor for flexureΦs: system factorΦv: resistance factor for shear
1© 2008 Taylor & Francis Group, LLC
1LRFD Method of Bridge Design
In load and resistance factor design (LRFD), bridges are designed for spe-cific limit states that consider various loads and resistance. These limit states include strength, extreme event, service, and fatigue, and are defined in the first section of this chapter. Subsequent sections cover the following load and resistance factors in more detail:
• Load combinations and load factors
• Strength limit states for superstructure design
• Resistance factors for strength limits
• Design live loads
• Number of design lanes
• Multiple presence of live loads
• Dynamic load allowances
• Live load distribution factors
• Load combinations for the Strength I Limit State
• Simple beam moments and shears carrying moving concentrated loads
Limit States
The following load combinations are defined in AASHTO. Bridges are designed for these limit states with consideration for the load and resistance factors detailed in later sections of this chapter.
A Art. 3.4.1, 1.3.2*
• Strength I: Basic load combination related to the normal vehicular use of the bridge without wind
• Strength II: Load combination relating to the use of the bridge by owner-specified special design vehicles and/or evaluation permit vehicles, without wind
* The article numbers in the 2010 Interim Revisions to the AASHTO Bridge Design Specifications, fifth edition, 2010 are by the letter A if specifications and letter C or Comm if commentary.
2 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
• Strength III: Load combination relating to the bridge exposed to wind velocity exceeding 55 mph without live loads
• Strength IV: Load combination relating to very high dead load to live load force effect ratios exceeding about 7.0 (e.g., for spans greater than 250 ft)
• Strength V: Load combination relating to normal vehicular use of the bridge with wind velocity of 55 mph
• Extreme Event I: Load combinations including earthquake• Extreme Event II: Load combinations relating to ice load or colli-
sions by vessels and vehicles• Service I: Load combination relating to the normal operational use
of the bridge with 55 mph wind. Also used for live load deflection control
Art. 2.5.2.6.2
• Service II: Load combination intended to control yielding of the steel structures and slip of slip-critical connections due to vehicular live load
• Service III: Load combination relating only to tension in prestressed concrete structures with the objective of crack control
• Fatigue I: Fatigue and fracture load combination related to infinite load-induced fatigue life
• Fatigue II: Fatigue and fracture load combination related to finite load-induced fatigue life
The following terms are defined for limit states:
γi = Load factor: a statistically based multiplier applied to force effectsϕ = Resistance factor: a statistically based multiplier applied to nomi-
nal resistance, as specified in AASHTO Specification Sections 5–8, and 10–12
ηi = Load modifier: a factor relating to ductility, redundancy, and oper-ational importance
ηD = A factor relating to ductility, as specified in AASHTO Article 1.3.3ηR = A factor relating to redundancy as specified in AASHTO Article
1.3.4ηl = A factor relating to operational importance as specified in
AASHTO Article 1.3.5Qi = Force effectRn = Nominal resistanceRr = Factored resistance: (ϕRn)
3LRFD Method of Bridge Design
© 2008 Taylor & Francis Group, LLC
Effects of loads must be less than or equal to the resistance of a member (or its components), or ηγQ ≤ ΦRn = Rr
A Art. 1.3.2
ΣηiγiQi ≤ ϕRn.
A Eq. 1.3.2.1-1
For loads for which a maximum value of γi is appropriate,
ηi = ηDηRηl ≥ 0.95
A Eq. 1.3.2.1-2
For loads for which a minimum value of γi is appropriate,
ηη η ηi
D R l= ≤1 1 0.
A Eq. 1.3.2.1-3
Load Combinations and Load Factors
Please see Tables 1.1 (A Tbl. 3.4.1-1) and 1.2 (A Tbl. 3.4.1-2), which show the load factors for various load combinations and permanent loads.
Loads and Load DesignationA Art. 3.3.2
The following permanent and transient loads and forces shall be considered in bridge design:
Permanent Loads
CR = force effects due to creep
DD = downdrag force
DC = dead load of structural components and nonstructural attachments
DW = dead load of wearing surfaces and utilities
EH = horizontal earth pressure load
4 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
TAB
LE 1
.1 (
AA
SH
TO
Tab
le 3
.4.1
-1)
Loa
d C
ombi
nati
ons
and
Loa
d F
acto
rs
Loa
d C
omb
inat
ion
L
imit
Sta
te
DC
DD
DW EH EV ES
EL
PS
CR
SH
LL
IM CE
BR
PL
LS
WA
WS
WL
FRT
UT
GSE
Use
On
e of
Th
ese
at a
Tim
e
EQ
ICC
TC
V
Stre
ngth
I (u
nles
s no
ted
)γp
1.75
1.00
——
1.00
0.50
/1.
20γ T
Gγ S
E—
——
—
Stre
ngth
IIγp
1.35
1.00
——
1.00
0.50
/1.
20γ T
Gγ S
E—
——
—St
reng
th II
Iγp
—1.
001.
40—
1.00
0.50
/1.
20γ T
Gγ S
E—
——
—St
reng
th IV
γp—
1.00
——
1.00
0.50
/1.
20––
––––
––––
––St
reng
th V
γp1.
351.
000.
401.
01.
000.
50/
1.20
γ TG
γ SE
——
——
Ext
rem
eE
vent
Iγp
γEQ
1.00
——
1.00
——
—1.
00—
——
Ext
rem
eE
vent
IIγp
0.50
1.00
——
1.00
——
——
1.00
1.00
1.00
Serv
ice
I1.
001.
001.
000
301.
01.
001.
00/
1.20
γ TG
γ SE
——
——
Serv
ice
II1.
001.
301.
00—
—1.
001.
00/
1.20
––––
––––
––––
Serv
ice
III
1.00
0.80
1.00
——
1.00
1.00
/1.
20γ T
Gγ S
E—
——
—Se
rvic
e IV
1.00
—1.
000.
70—
1.00
1.00
/1.
20—
1.0
——
——
Fati
gue
I—LL
, IM
, & C
E o
nly
—1.
50—
——
——
——
——
——
Fati
gue
II—
LL, I
M, &
CE
onl
y—
0.75
——
——
——
——
——
—
5LRFD Method of Bridge Design
© 2008 Taylor & Francis Group, LLC
EL = miscellaneous locked-in force effects resulting from the con-struction process, including jacking apart of cantilevers in seg-mental construction
ES = earth surcharge loadEV = vertical pressure from dead load of earth fillPS = secondary forces from posttensioningSH = force effects due to shrinkageγp = load factor for permanent loading
Transient Loads
BR = vehicular braking forceCE = vehicular centrifugal forceCT = vehicular collision forceCV = vessel collision forceEQ = earthquake loadFR = friction load
TABLE 1.2 (AASHTO Table 3.4.1-2)
Load Factors for Permanent Loads, γp
Type of Load. Foundation Type, and Method Used to Calculate Downdrag
Load Factor
Maximum Minimum
DC: Component and Attachments 1.25 0.90DC: Strength IV only 1.50 0.90DD: Downdrag Piles, α Tomlinson Method 1.4 0.25
Piles. λ Method 1.05 0.30Drilled shafts. O’Neill and Reese (1999) Method 1.25 0.35
DW: Wearing Surfaces and Utilities 1.50 0.65EH: Horizontal Earth Pressure•Active 1.50 0.90•At-Rest 1.35 0.90•AEP for Anchored Walls 1.35 N/A
EL: Locked-in Construction Stresses 1.00 1.00EV: Vertical Earth Pressure•OverallStability 1.00 N/A•RetainingWallsandAbutments 1.35 1.00•RigidBuriedStructure 1.30 0.90•RigidFrames 1.35 0.90
ES: Earth Surcharge 1.50 0.75
6 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
IC = ice loadIM = vehicular dynamic load allowanceLL = vehicular live loadLN = design lane loadLS = live load surchargePL = pedestrian live loadSE = force effect due to settlementTG = force effect due to temperature gradientTL = design truck load or design tandem loadTU = force effect due to uniform temperatureWA = water load and stream pressureWL = wind on live loadWS = wind load on structure
Strength Limit States for Superstructure Design
The load effect, Q, is given in the following equations in relation to various limit states:
Strength I: Q = 1.25 DC + 1.5 DW + 1.75 LL; LL = TL + LNStrength II: Q = 1.25 DC + 1.5 DW + 1.35 LLStrength III: Q = 1.25 DC + 1.5 DW + 1.4 WSStrength IV: Q = 1.5 DC + 1.5 DWService I: Q = 1.0 DC + 1.0 DW + 1.0 LL + 1.0 WA + 0.3 WS + 1.0 WLService II: Q = 1.0 DC + 1.0 DW + 1.3 LLService III: Q = 1.0 DC + 1.0 DW + 0.8 LL +1.0 WAFatigue I: Q = 1.5 (LL + IM)Fatigue II: Q = 0.75 (LL + IM)
Resistance Factors, Φ, for Strength Limits
Resistance factors, Φ, for strength limit states are given for various structural categories in AASHTO. Selected resistance factors that are most frequently encountered follow.
7LRFD Method of Bridge Design
© 2008 Taylor & Francis Group, LLC
For service and extreme event limit states, resistance factors ϕ shall be taken as 1.0, except for bolts.
A Art. 1.3.2.1
Flexure and tension in reinforced concrete : 0.90A Art. 5.5.4.2
Flexure and tension in prestressed concrete : 1.00Shear in concrete : 0.90Axial compression in concrete : 0.75Flexure in structural steel : 1.00
A Art. 6.5.4.2
Shear in structural steel : 1.00Axial compression in structural steel : 0.90Tension, yielding in gross section : 0.95
Design Live Load HL-93HL-93 is a notional live load where H represents the HS truck, L the lane load, and 93 the year in which the design live load HL-93 was adopted.
The design live load designated as the HL-93 consists of a combination of:A Art. 3.6.1.2
• Design truck (HS-20) or design tandem (a pair of 25 kip axles 4 ft apart),• Design lane load of 0.64 kip-ft uniformly distributed in the longitu-
dinal direction. Transversely the design lane load is distributed over a 10 ft width within a 12 ft design lane. Note that the design lane load is not subject to dynamic allowance.
For both design truck and design tandem loads, the transverse spacing of wheels is taken as 6.0 ft. The uniform lane load may be continuous or discon-tinuous as necessary to produce the maximum force effect.
Fatigue Live Load The fatigue load will be one truck or axles as specified in Figure 1.1 ( A Art. 3.6.1.2.2 ) with a constant spacing of 30 ft between the 32.0 kip axles. The frequency of the fatigue load shall be taken as the single-lane average daily truck traffic. When the bridge is analyzed by approximate load distribution (A Art. 4.6.3), the distribution factor for one traffic lane shall be used. See Figure 1.2.
A Art. 3.6.1.4
8 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
25 kips
Designtandem
0.64 kips/ftDesign lane
0.64 kips/ftDesign lane
2 ft 2 ft6 ft
0.64 kips/ft
12 ft lane width
Live load placement
Section A-A
4 ft
25 kips32 kips 32 kips
14 ft to 30 ft14 ftA
A
8 kips
Design truck
FIGURE 1.1Design truck (HS-20), design tandem load (a pair of 25 kip axles 4 ft apart), and design lane load (0.64 kips/ft longitudinally distributed). (Source: Art. 3.6.1.2.2. )
32 kips 32 kips
30 ft14 ft
8 kips
FIGURE 1.2Fatigue live loading. (Source: Art. 3.6.1.4. )
9LRFD Method of Bridge Design
© 2008 Taylor & Francis Group, LLC
Number of Design Lanes, NL
The number of design lanes, NL, is the integer portion of the ratio of the clear roadway width (ft), w, and the width of the design traffic lane (12 ft). For example, if the clear roadway width is 40 ft,
A Art. 3.6.1.1.1
N wft
lane
ftft
lane
lanes lL = = =12 0
40
12 03 33 3
. .. ( aanes)
Multiple Presence Factor of Live Load, m
Design trucks will be present in adjacent lanes on roadways with multiple design lanes. Because it is unlikely that three or more adjacent lanes will be loaded simultaneously with trucks, adjustments in design loads are necessary. These factors have been implicitly included in the approximate equations for distribu-tion factors and should be removed for fatigue investigations. See Table 1.3.
A Art 3.6.1.1.2
Therefore for fatigue investigations in which the traffic truck is placed in a single lane, the factor of 1.2 which has been included in the approximate equations should be removed.
A Com. 3.6.1.1.2
The multiple presence m is defined for sites with an ADTT of 5,000 trucks or greater in one direction. For sites with a lower ADTT, reduce force effects by:
TABLE 1.3 (AASHTO Table 3.6.1.1.2-1)
Multiple Presence Factors, m
A Art. 3.6.1.1.2
Number of Loaded Lanes
Multiple Presence Factors, m
1 1.202 1.003 0.85
>3 0.65
10 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
100 < ADTT < 1,000 → 95%
ADTT < 100 → 90%
This adjustment is based on the reduced probability of attaining the design event during a 75-year design life with reduced truck volume.
Dynamic Load Allowance, IM
The dynamic load allowance, IM, is applied only to the design truck load, or tandem, not to the design lane load. The static effects of the design truck or tandem shall be increased for dynamic load allowance. Table 1.4 indicates the dynamic load allowance for different components under different limit states.
A Art. 3.6.2
Live Load Distribution Factors
Live load distribution factors in AASHTO are lane-load distributions, not wheel-load distributions as they were in the AASHTO Standard Specifications for Highway Bridges, 17th edition, 2002.
A Art. 4.6.2.2: Appendices A–D
The distribution factors are included in several AASHTO articles, and important provisions are in AASHTO Sec. 4.
The live load moment and shear for beams or girders are determined by applying the lane fraction (distribution factor) in AASHTO Art. 4.6.2.2.2 to the moment and shear due to the loads assumed to occupy 10 ft. transversely within a design lane.
AASHTO 3.6.1.2.1
TABLE 1.4 (AASHTO Table 3.6.2.1-1)
Dynamic Load Allowance, IM
Component IM (%)
Deck joints, all limit states 75All other components Fatigue and fracture limit state 15 All other limit states 33
11LRFD Method of Bridge Design
© 2008 Taylor & Francis Group, LLC
Distribution factors are most sensitive to beam (girder) spacing. Span length and longitudinal stiffness have smaller influences.
Load-carrying capacity of exterior beams (girders) shall not be less than that of interior beams (girders).
A Art. 2.5.2.7.1
Because of the many algebraically complex expressions and equations associated with live load distribution, they are discussed in detail in exam-ple problems.
Load Combinations for the Strength I Limit State
The total factored force effects, Q, shall be taken asA Arts 3.3.2, 3.4.1, Tbls. 3.4.1-1, 3.4.1-2
Q = ΣηiγiQi
Where:
ηi = load modifier = 1.0γi = load factor
A Eq. 3.4.1-1
For load combination limit state Strength I,
Q = 1.25 DC + 1.50 DW + 1.75 (TL + LN)Mu = 1.25 MDC +1.50 MDW + 1.75 (MTL + MLN)Vu = 1.25 VDC +1.50 VDW + 1.75 (VTL + VLN)TL = Truck load or tandem loadLN = Lane loadηD = 1.0 for conventional designs
A Art. 1.3
ηR = 1.0 for conventional levels of redundancyηI = 1.0 for typical bridgesΦ = 1.0 for service and fatigue limit states
A 13.2.1; A 5.5.4.2; A 6.5.4.2; A C6.6.1.2.2
ϕ For concrete structures and steel structures, refer to Art. 5.5.4.2.1 and Art. 6.5.4.2 , respectively.
12 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Unfactored Dead Load Moments and Shears
MDC = Maximum unfactored moment due to DCMDW = Maximum unfactored moment due to DWVDC = Maximum unfactored shear due to DCVDW = Maximum unfactored shear due to DW
Unfactored Live Load Moments per beam with Distribution Factors DFM, and Dynamic Allowance IM
MTL = (maximum truck or tandem load moment per lane due to design truck load) (DFM) (1 + IM)
MLN = (maximum lane load moment per lane) (DFM)*
Unfactored Live Load Shear per Beam with Distribution Factors DFV, and Dynamic Allowance, IM
VTL = (maximum truck load per shear lane) (DFV) (1+IM)VLN = (maximum lane load shear per lane) (DFV)2
Live Load Distribution Factors for Moment for Beams
DFMsi = Single (one) lane loaded for moment in interior beamsDFMmi = Multiple (two or more) lanes loaded for moment in interior
beamsDFMse = Single (one) lane loaded for moment in exterior beamsDFMme = Multiple (two or more) lanes loaded for moment in exterior
beams
Live Load Distribution Factors for Shear for Beams
DFVsi = Single (one) lane loaded for shear in interior beamsDFVmi = Multiple (two or more) lanes loaded for shear in interior
beamsDFVse = Single (one) lane loaded for shear in exterior beamsDFVme = Multiple (two or more) lanes loaded for shear in exterior
beams
* No impact allowance applies.
13LRFD Method of Bridge Design
© 2008 Taylor & Francis Group, LLC
Simple Beam Live Load Moments and Shears Carrying Moving Concentrated Loads per Lane
The maximum moment occurs under one of the loads when that load is as far from one support as the center of gravity of all the moving loads on the beam is from the other support. This condition occurs when the center of the span is midway between the center of gravity of the moving loads and the nearest concentrated load where the maximum moment occurs.
The maximum shear due to moving loads occurs at one support when one of the moving loads is at that support. With several moving loads, the location that will produce maximum shear must be determined by trial and error, as shown. Because the maximum moments for uniform loads such as the lane loads and dead loads occur at midspan, the maximum design truck or tandem moment is generally used with the HL-93 center axle at midspan. See Figures 1.3 and 1.4.
Live Load Moments and Shears for Beams (Girders)
A Art. Tbls. 4.6.2.2.2b-1 and 4.6.2.2.2d-1; and 4.6.2.2.3a-1 and 4.6.2.2.3b-1
Live load moments and shears for beams (girders) are determined by apply-ing the distribution factors for live loads per lane in AASHTO tables in 4.4.2.2 to the moment and shears due to the live loads assumed to occupy 10 ft. within a design lane (AASHTO 3.6.1.2.1).
14 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
32.64 kips
7.36 kips
25 ft
2 ft 4 in2 ft 4 in
13 ft 4 in
32 kips 32 kips 8 kips
c.g.of truck
25 ft 32.64 kips
8 ft 8 in14 ft9 ft 4 in
39.36 kips
Span
Span
627.8 ft-kips
24.64 kips
7.36 kipsShear(kips)
Moment(ft-kips)
39.36 kips
CL
CL
FIGURE 1.3Shear and moment diagrams for controlling design truck (HS-20) live load position.
15
© 2008 Taylor & Francis Group, LLC
29.28 kips
10.72 kips
25 ft
14 ft4 ft8 in
11 ft
32 kips 32 kips 8 kips
c.g.of truck
25 ft 29.28 kips
11 ft
9 ft 4 in
42.72 kipsSpan
Span
620 ft-kips
21.28 kips
Shear(kips)
Moment(ft-kips)
42.72 kips
CL
CL
FIGURE 1.4Shear and moment diagrams for the design truck (HS-20) center axle at midspan.
17© 2008 Taylor & Francis Group, LLC
2Design Examples
Design Example 1: Reinforced Concrete T-Beam Bridge
Problem Statement
A bridge will be designed with a span length of 50 ft. The superstructure consists of five beams spaced at 10 ft with a concrete deck slab of 9 in. The overall width of the bridge is 48 ft and the clear (roadway) width is 44 ft, 6 in. Design the superstructure of a reinforced, cast-in-place concrete T-beam bridge using the following design specifications.
The three Load Combination Limit States considered are Strength I, Fatigue II, and Service I.
C&P Curb and parapet cross section 3.37 ft2
Ec Modulus of elasticity of concrete 4 × 103 kips/in2
f′c Specified compressive strength of concrete 4.5 kips/in2
fy Specified yield strength of epoxy-coated 60 kips/in2
reinforcing barswc Self-weight of concrete 0.15 kips/ft3
wFWS Future wearing surface load 0.03 kips/ft2
Figure 2.1 shows the elevation view, section view, and overhang detail of the reinforced concrete T-beam bridge described
Solution
Step 1: Design T-Beam Using Strength I Limit State
The factored load, Q, is calculated using the load factors given in AASHTO Tables 3.4.1-1 and 3.4.1-2.
Q = 1.25 DC + 1.5 DW + 1.75(TL+LN)
18 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Find the flange width and web thickness.A Art. 5.14.1.5.1
For the top flange of T-beams serving as deck slabs, the minimum concrete deck must be greater than or equal to 7 in. First, try using a slab with a thick-ness, ts, of 9 in.
A Arts. 5.14.1.5.1a, 9.7.1.1
The minimum web thickness is 8 in.A Art. 5.14.1.5.1c; Com. 5.14.1.5.1c
Minimum concrete cover for main epoxy-coated bars shall be 1 in. Use 1.5 in cover for main and stirrup bars.
A Tbl. 5.12.3-1; Art. 5.12.4
Find the width of T-beam stem (web thickness), b.A Art. 5.10.3.1.1
The minimum width, bmin, is found as follows:
Assume two layers of no. 11 bars for positive reinforcement.
db for a no. 11 bar = 1.41 in
Four no. 11 bars in a row and no. 4 stirrup bars require a width of
bmin = 2 (2.0 in cover and no. 4 stirrup) + 4 db + 3 (1.5 db)
= 2 (2.0 in) + 4 (1.41 in) + 3 (1.5 × 1.41 in) = 16 in
24 ft24 ft
10 ft 10 ft 10 ft
Curb andparapet
4 ft 4 ft10 in
4 ft
x = 0.66 ft
18 inOverhang detail
3 in9 in slab
1 ft 7 in
7 in2 in
12 in
S = 10 ft
2 ft 8 in
1 ft 9 in 44 ft 6 in
Elevation
L = 50 ftA
A
1 ft 9 in
ts
Section A-A
–
FIGURE 2.1T-beam design example.
19Design Examples
© 2008 Taylor & Francis Group, LLC
Try bw = width of stem = 18 in.
Find the beam depth including deck.A Tbl. 2.5.2.6.3-1
hmin = 0.070 L = 0.070 (50 ft × 12 in) = 42 in.
Try h = beam width = 44 in.
Find the effective flange width.A Art. 4.6.2.6.1
Find the effective flange width, where
be effective flange for exterior beams inbi effective flange width for interior beams inbw web width 18 inL effective span length (actual span length) 50 ftS average spacing of adjacent beams 10 ftts slab thickness 9 in
The effective flange width for interior beams is equal to one-half the dis-tance to the adjacent girder on each side of the girder,
bi = S = 10 ft × 12 in = 120 in
The effective flange width for exterior beams is equal to half of one-half the distance to the adjacent girder plus the full overhang width,
be = ½ (10 ft × 12 in) + (4 ft × 12 in) = 108 in
Find the interior T-beam section. Please see Figure 2.2.
The section properties of the preceding interior T-beam are as follows.
The area of the T-beam is
A = (9 in) (120 in) + (35 in) (18 in) = 1710 in2
The center of gravity from the extreme tension fiber is
y y AA
in in in in int
t= ∑∑
= + +( )( )( . ) ( )(9 120 35 4 5 35 188 17 51710 0 2
in inin
)( . ).
= 31.4 in
20 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
The moment of inertia (the gross concrete section) about the center of grav-ity is
I I I Ad
in in in in
g = = ∑ +( )= +
2
3120 912
120 9( )( ) ( )( )(112 5 4 5
18 3512
18 35
2
3
. . )
( )( ) ( )(
in in
in in in in
−
+ + ))( . . )
, .
31 4 17 5
264 183 6
2
4
in in
I ing
−
=
The T-beam stem is
(35 in)(18 in) = 630 in2
The section modulus at bottom fiber is
SIy
inin
ing
t= = =264 183 6
31 48413 5
43, .
..
Find the number of design lanes.
A Art. 3.6.1.1.1
The number of design lanes, NL, is the integer portion of the ratio of the clear road width divided by a 12 ft traffic lane width.
18 in
be = 120 in
c.g.
12.6 in
9 in
35 in yt = 31.4 in
FIGURE 2.2Interior T-beam section.
21Design Examples
© 2008 Taylor & Francis Group, LLC
N wft
lane
ftft
N lanes
L
L
= =
=
12
44 512
3 7 3
.
. ( )
Find the truck load moments and shears.
Design truck load (HS-20) for moment at midspan (Figure 2.3a).Design tandem load for moment at midspan (Figure 2.3b).The design truck load (HS-20) is shown in Figure 2.4. Figures 2.5 and 2.6
show the design tandem load position for moment at midspan and the design truck load (HS-20) position for shear at support, respectively. The design tan-dem load position for shear at support is shown in Figure 2.7.
NOT E: The maximum design truck or design tandem moment is generally used with the HL-93 center axle at midspan.
25 ft
11 ft4 ft11 ft
5.5 5.5
10.512.5
25 ft
FIGURE 2.3aInfluence lines for moment at midspan.
10 ft4 ft
1.0 0.920.72
0.44
14 ft 22 ft
FIGURE 2.3bInfluence lines for shear at support.
22 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
The truck load moment per lane is
Mtr = 32 kips (12.5 ft + 5.5 ft) + 8 kips (5.5 ft)
= 620 kip-ft
The tandem load moment per lane is
Mtandem = 25 kips (12.5 ft + 10.5 ft)
= 575 kip-ft
11 ft 14 ft
32 kips 32 kips 8 kips
14 ft 11 ft
CL
FIGURE 2.4Design truck (HS-20) position for moment at midspan.
4 ft
25 kips25 kips
25 ftCL
FIGURE 2.5Design tandem load position for moment at midspan.
14 ft 22 ft
32 kips 8 kips
14 ft
32 kips
FIGURE 2.6Design truck (HS-20) position for shear at support.
46 ft4 ft
25 kips 25 kips
FIGURE 2.7Design tandem load position for shear at support.
23Design Examples
© 2008 Taylor & Francis Group, LLC
The truck load shear per lane is
Vtr = 32 kips (1 + 0.72) + 8 kips (0.44)
= 58.6 kips
The tandem shear per lane is
Vtandem = 25 kips (1 + 0.92)
= 48 kips
The lane load per lane is
Mln = =
=wL
kipsft
ftkip ft
22
8
0 64 50
8200
. ( )-
The lane load shear per lane is
Vln = =
=wL
kipsft
ftkips
2
0 64 50
216
. ( )
Find the live load distribution factors for moments, DFM.
For cast-in-place concrete T-beam, the deck type is (e).A Tbl. 4.6.2.2.1-1
For interior beams,A Tbl. 4.6.2.2.2; A Art. 4.6.2.2.2b; Tbl. 4.6.2.2.2b-1; Appendix A*
Kg longitudinal stiffness in4
L span length of beams 50 ftNb number of beams 5S spacing of beams 10 ftts slab thickness 9 in3.5 ft ≤ S ≤ 16 ft S = 10 ft OK
4.5 in ≤ ts ≤ 12 in ts = 9 in OK
20 ft ≤ L ≤ 240 ft L = 50 ft OK
Nb ≥ 4 Nb = 5 OK
* Refer to the Appendices at the end of the book.
24 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
The modular ratio between beam and deck material, n, is 1.0.A Eq. 4.6.2.2.1-2
A area of beam or T-beam 1710 in2
L span length 50 ftts slab thickness 9 inn modular ratio between beam and deck material 1
The moment of inertia of the basic beam (portion of beam below deck) is
I = =( )( ) , .18 3512
64 312 53
4in in in
The distance between the centers of gravity of the basic beam and deck is
eg = 17.5 in + 4.5 in = 22 in
The longitudinal stiffness parameter isA Eq. 4.6.2.2.1-1
Kg = n [I + (A)(eg2)] = 1[64312.5 in4 + (1,710 in2)(22 in)2]
Kg = 891,953.0 in4
A simplified value may be considered.A Tbl. 4.6.2.2.1-2
KLt
g
s121 053
0 1
=
.
.
Multiple presence factors, m, shall not be applied in conjunction with approximate load distribution factors specified in Art. 4.6.2.2 and 4.6.2.3, except where the lever rule is used.
A Art. 3.6.1.1.2
The distribution factor for moment for interior beams with one design lane, where si is the single lane loaded in interior beams is found as follows.
A Tbl. 4.6.2.2.2b-1 or Appendix A
The multiple presence factor, m, is applicable only when the lever rule is used for the distribution factors. Therefore, m = 1.0 where DFM is the distri-bution factor for moment.
25Design Examples
© 2008 Taylor & Francis Group, LLC
DFM m S SL
KLtsi
g
s= +
0 0614 12
0 4 0 3
3.. .
= +
0 1
0
1 0 0 06 1014
.
.
( . ) . ft 44 0 31050
1 05
0 629
ftft
lan
( )
=
.
.
. e//girder
The distribution factor for moment for interior beams with two or more design lanes loaded, where mi is the multiple lanes loaded in interior beams is
A Tbl. 4.6.2.2.2b-1 or Appendix A
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
0 6
0 075 109 5
1
= +
.
.
..ft 00
501 05
0 859
0 2ftft
lane
( )
=
.
.
. /giirder governs for interior beams
For the distribution of moment for exterior beams with one design lane, use the lever rule. See Figure 2.8.
A Art. 3.6.1.3.1, A Art. 4.6.2.2.2d; Tbl. 4.6.2.2.2d-1 or Appendix B
4 ft 10 ftde = 2.25 ft
5.75 ft
6 ft2 ft
1.75ft
4.25 ft
P2
18 inR
P2
Assume hinge
d
FIGURE 2.8Lever rule for determination of distribution factor for moment in exterior beam, one lane loaded.
26 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Σ Moment at “a” = 0
0 102
10 252
4 25
0 725
= − ( ) + ( ) + ( )=
R ft P ft P ft
R P
. .
.
The multiple presence factor for one design lane loaded, m, is 1.20.A Tbl. 3.6.1.1.2-1
The distribution factor for moment for exterior beams for one design lane loaded is, where se is the designation for single lane loaded in the extreme beam,
DFMse = m[0.725] = 1.20 [0.725]
= 0.87 lane/girder [governs for exterior beam]
The exterior web of the exterior beam to the interior edge of the curb, de, is 2.25 ft, which is OK for the –1.0 ft ≤ de ≤ 5.5 ft range.
A Tbl. 4.6.2.2.2d-1 or Appendix B
The distribution factor for the exterior beam is
e g e DFMmi( )( ) = ( )( )interior
e de= +0 779 1
..
e = +0 77 2 259 1
. ..
e = 1.017
Use e = 1.0.
The distribution factor for moment for exterior beams with two or more design lanes loaded, where me is the designation for multiple lanes loaded in the exterior beam, is
A Tbl. 3.6.1.1.2-1
DFMme = (m)(e)(ginterior) = (m)(e)(DFMmi) = (0.85)(1.0)(0.859)
= 0.730 lane/girder
27Design Examples
© 2008 Taylor & Francis Group, LLC
Find the distributed live load moments.
The governing distribution factors are:
Interior beam DFMmi = 0.859 lane/girder
Exterior beam DFMse = 0.870 lane/girder
IM = 33%
A Tbl. 3.6.2.1-1
The unfactored live load moment per beam for interior beams due to truck load is
MTL = Mtr(DFM)(1 + IM) = (620 ft-kips)(0.859)(1+0.33)
= 708.33 ft-kips
The unfactored live load moment per beam for interior beams due to lane load is
MLN = Mln (DFM) = (200 ft-kips)(0.859)
= 171.8 ft-kips
The unfactored live load moment per beam for exterior beams due to truck load is
MTL = Mtr(DFM)(1 + IM) = (620 ft-kips)(0.87)(1 + 0.33)
= 717.40 ft-kips
The unfactored live load moment per beam for exterior beams due to lane load is
MLN = Mln (DFM) = (200 ft-kips)(0.87)
= 174.0 ft-kips
Find the distribution factors for shears, DFV.
A Art. 4.6.2.2.3
For a cast-in-place concrete T-beam, the deck type is (e).
A Tbl. 4.6.2.2.1-1
28 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
The distribution factor for shear for interior beams with one design lane loaded is
A Tbl. 4.6.2.2.3a-1 or Appendix C
DFV S ft
si = +
= +
=
0 3625
0 36 1025
0 76
. .
. llanes
The distribution factor for shear for interior beams with two or more design lanes loaded is
DFV S S ftmi = + −
= +0 212 35
0 2 102 0
. ..
1121035
2 0
−
ft .
DFVmi = 0.95 lane/girder [controls for interior beams]
Using the lever rule for moment, the distribution factor for shear for exte-rior beams with one design lane loaded is
A Tbl. 4.6.2.2.3b-1
DFVse = DFMse = 0.87 lane/girder [controls for exterior beams]
The distribution factor for shear for exterior beams with two or more design lanes loaded is
g = (e)ginterior = (e)(DFVmi)
e d fte= + = + =0 610
0 6 2 2510
0 825. . . .
DFVme = mg = (m)(e)(ginterior), where ginterior = DFVmi
A Tbl. 3.6.1.1.2-1 = (1.0)(0.825)(0.95)
= 0.784 lane/girder
Find the distributed live load shears for Strength I.
The governing distribution factors are:
Interior beam DFVmi = 0.95 lane/girderExterior beam DFVse = 0.87 lane/girder
29Design Examples
© 2008 Taylor & Francis Group, LLC
The unfactored live load shears per beam due to truck load for interior beams is VTL = Vtr(DFV)(1 + IM)
= (58.6 kips)(0.95)(1.33)
= 74.04 kips
The unfactored live load shear per beam due to lane load for interior beams is
VLN = Vln(DFV)
= (16.0 kips)(0.95)
= 15.2 kips
The unfactored live load shears per beam for exterior beams are
VTL = Vtr(DFV)(1 + IM)
= (58.6 kips)(0.87)(1.33)
= 67.80 kips
VLN = Vln(DFV)
= (16.0 kips)(0.87)
= 13.92 kips
Find the dead load force effects.
The self-weights of the T-beam, the deck, and the curb and parapet for interior beams are represented by the variable DC.
DC in ftin
kipsfT beam− = ( )
1710 1144
0 1522
2 .tt3
= 1 78. kipsft
DC ft kipsft beamsC P& . .= ( )
2 3 37 0 15 15
23
= 0 202. kipsft
30 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
wDC = DCT-beam + DCC&P
= +1 78 0 202. .kipsft
kipsft
= 1 98. kipsft
The corresponding shear is
V wLkips
ftft
DC = =
( )
2
1 98 50
2
.
= 49.5 kips
The corresponding moment is
M wLkips
ftft
DC = =
( )2
2
8
1 98 50
8
.
= 618.75 kip-ft
The future wearing surface load, DW, per beam is
w kipsft
ftbeamDW =
0 03 102.
= 0 3. kipsft
The corresponding shear is
V w Lkips
ftft
DWDW= =
( )
2
0 3 50
2
.
= 7.5 kips
31Design Examples
© 2008 Taylor & Francis Group, LLC
The corresponding moment is
M w Lkips
ftft
DWDW= =
( )2
2
8
0 3 50
8
.
= 93.75 kip-ft
The unfactored interior beam moments and shears due to the dead loads plus the live loads are given in Table 2.1. Note that dead loads consist of the exterior T-beam stem, deck slab, curb/parapet, and wearing surface.
For exterior girders, the deck slab load is,
w in ftin
kipsfts = ( )
9 112
0 15 3.
= 0 113 2. kipsft
For exterior girders, the wearing surface load is given as
w kipsftDW = 0 03 2.
The total load is
w w w
kipsft
kipsft
kips
S DW= +
= +
=
0 113 0 03
0 143
2 2. .
.fft2
See Figure 2.9.
TABLE 2.1
Distributed Live Load and Dead Load Effects for Interior Beam for Reinforced Concrete T-Beam Bridge
Moment M(ft-kips)
Shear V(kips)
DC 618.75 49.5DW 93.75 7.5TL 708.33 74.04LN 171.8 15.2
32 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Σ moments about B = 0
0 = –0.952 ft-kips – RA(10 ft) + (0.143 kips/ft2)(14 ft)(7 ft).
RA = 1.31 kips per foot of exterior beam due to the deck slab and wear-ing surface dead loads
The load for the curb and parapet for exterior girders is,
w ft kipsftC P& . .= ( )
3 37 0 1523
= 0 51. kipsft
See Figure 2.10.
Σ moments about B = 0
0 = 0.476 ft-kips – RA(10 ft) + (0.51 kips)(13.34 ft)
RA = 0.728 kips per foot of exterior beam due to the curb and parapet dead loads
4 ft4 ft 10 ft10 ft
–0.94
symm.
–0.952–0.904+0.904 +0.94+0.952
10 ft
B
w = 0.143 kips/ft2
0.143 kips/ft2
0.952kips/ft10 ft4 ft
RA RB
BA
A
10 ft
By the momentdistribution methodM, ft-kips
CL
FIGURE 2.9Moment distribution for deck slab and wearing surface loads.
33Design Examples
© 2008 Taylor & Francis Group, LLC
The reactions at exterior beam due to the dead loads are as follows
DC deck slab =
1 310 113
0 143
2
2
..
.
kipsft
kipsft
kipsft
= 1 04. kipsft
Curb and parapet overhang
w kipsftC P& .= 0 728
girder stem = ( )
=630 1144
0 1522
2 3in ftin
kipsft
. 00 656. kipsft
wDC = + + =1 04 0 728 0 656 2 42. . . .kipsft
kipsft
kipsft
kipssft
By the momentdistribution methodM, ft-kips
4 ft
0.66 ft
+0.476
0.476kips/ft symm.
CL
–1.67+1.67 –0.476
10 ft
BA
10 ft
WC and P = 0.51 kips
10 ft
3.34
ft
3.34ft
0.51 kips
4 ft
RA RB
B
FIGURE 2.10Moment distribution for curb and parapet loads for exterior girder.
34 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Future wearing surface
w kipsft
kipsftkipsft
DW =
1 310 03
0 143
2
2
..
.
= 0 27. kipsft
The unfactored exterior girder moments and shears due to the dead loads plus the live loads are as follows.
MDC = =
( )
=w Lkips
ftft
ki fDC2
2
8
2 42 50
8756 3
.. p- tt
VDC = =
( )
=w Lkips
ftft
kipsDC
2
2 42 50
260 5
..
MDW = =
( )
=w Lkips
ftft
kDW2
2
8
0 27 50
884 4
.. ip-ft
VDW = =
( )
=w Lkips
ftft
kipsDW
2
0 27 50
26 75
..
The unfactored beam moments and shears due to the dead loads plus the live loads are given in Table 2.2.
TABLE 2.2
Unfactored Beam Moments and Shears Due to Dead Loads and Live Loads for Reinforced Concrete T-Beam Bridge
Interior Beam Exterior Beam
Moment M (ft-kips)
Shear V(kips)
Moment M (ft-kips)
Shear V(kips)
DC 618.75 49.5 756.3 60.5DW 93.75 7.5 84.4 6.75TL 708.33 74.04 717.4 67.8LN 171.8 15.2 174.0 13.92
35Design Examples
© 2008 Taylor & Francis Group, LLC
Find the factored moments and shears for Strength I.
A Tbls. 3.4.1-1; 3.4.1-2]
Q = 1.25 DC + 1.5 DW + 1.75(TL + LN)
For interior girders, the unfactored moment is
Mu = 1.25 MDC + 1.5 MDW + 1.75 (MTL + MLN)
= 1.25 (618.75 kip-ft) + 1.5 (93.75 kip-ft) + 1.75 (708.33 kip-ft + 171.8 kip-ft)
= 2454.29 kip-ft
For interior girders, the factored shear is
Vu = 1.25 VDC + 1.5 VDW + 1.75 (VTL + VLN)
= 1.25 (49.5 kips) + 1.5 (7.5 kips) + 1.75 (74.04 kips+ 15.2 kips)
= 229.2 kips [controls]
For exterior girders, the factored moment is
Mu = 1.25 MDC + 1.5 MDW + 1.75 (MTL + MLN)
= 1.25 (756.3 kip-ft) + 1.5 (84.4 kip-ft) + 1.75 (717.4 kip-ft + 174.0 kip-ft)
= 2631.9 kip-ft [controls]
For exterior girders, the factored shear is
Vu = 1.25 VDC + 1.5 VDW + 1.75 (VTL + VLN)
= 1.25 (60.5 kips) + 1.5 (6.75 kips) + 1.75 (67.8 kips+ 13.92 kips)
= 228.8 kips
Find the design flexural reinforcements, neglecting compression rein-forcement, and note the exterior girder moment and interior girder shear control.
Mu = 2631.9 kip-ft
Vu = 229.2 kips
36 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
For stem reinforcement, try 12 no. 11 bars. See Figure 2.11.Web thickness, bw, is,
A Art. 5.14.1.5.1c
bw = 2(1 in) + 2(0.5 in) + 4 db + 3(1.5 db)
= 2 in + 1 in + 4(1.4 in) + 3(1.5 x 1.4 in)
= 14.9 in
No. 4
18 in
T-Beam Section
be = 120 in
c.g.
3.6 in
5.5 in
h = 44 inds
9 in
35 in yt = 31.4 in
18 in
1.41 in
4.19 in
No. 11
Arrangement of barsin stem of T-beam
2.78 in
5.5 in
1 in
1.5 inclearance
FIGURE 2.11T-beam section and reinforcement in T-beam stem.
37Design Examples
© 2008 Taylor & Francis Group, LLC
To give a little extra room, use bw = 18 inThe distance from extreme compression fiber to the centroid of tensile
reinforcement is
ds = 44 in – 5.5 in
= 38.5 in
db = 1.41 in [no. 11 bar]
For 12 no. 11 bars, As = 18.72 in2.The minimum clearance between bars in a layer must not be less than
A Art. 5.10.3.1.1
1.5 db = 1.5 (1.41 in) = 2.1 in
2.1 in < provided = 2.78 in [OK]
The clear distance between layers shall not be less than 1 in or db.
A Art. 5.10.3.1.3
db = 1.41 in [OK]
Concrete cover for epoxy-coated main reinforcing bars is 1.0 in, provided = 1.5 in [OK]
A Art. 5.12.3
Φ = 0.90
A Art. 5.5.4.2
As = 18.75 in2
fy = 60.0 ksi
ds = 38.5 in
The thickness of the deck slab, ts, is 9 in and the width of compression face, b, is 120 in.
The factor for concrete strength β1 is
A Art. 5.7.2.2
38 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
β1 = −′ −
0 854000
10000
2
2
.f kips
inkipsin
c..05( )
= −−
0 854500 4000
1000
2 2
2
.kipsin
kipsin
kipsin
( )0 05.
= 0.825
The distance from the extreme compression fiber to the neutral axis, c, isA Eq. 5.7.3.1.1-4
c =′
A ff bs y
c0 85 1. β
=( )
18 75 60
0 85 4 5
22
2
.
. .
in kipsin
kipsin
( )( )0 825 120. in
c = 2.97 in
a = c β1
= (2.97 in)(0.825)
= 2.45 in < ts = 9 in [OK]
The nominal resisting moment, Mn, isA Art. 5.7.3.2.2
Mn = −
A f d as y s 2
= ( )
−
18 75 60 38 5 2 452
22. . .in kips
inin in
112
ftin
= 3488.9 in2
39Design Examples
© 2008 Taylor & Francis Group, LLC
The factored resisting moment, Mr, isA Art. 5.7.3.2.1
Mr = ΦMn
= 0.90(3488.9 ft-kips)
= 3140.0 ft-kips > Mu = 2631.9 ft-kips [OK]
Check the reinforcement requirements.
A Art. 5.7.3.3.1; 5.7.3.3.2
This provision for finding the maximum reinforcement was deleted from AASHTO in 2005. Therefore, check for the minimum reinforcement. The minimum reinforcement requirement is satisfied if Mr is at least equal to the lesser of:
The minimum reinforcement requirement is satisfied if ΦMn (= Mr) is at least equal to the lesser of:
• Mr ≥ 1.2 Mcr
• Mr ≥ 1.33 times the factored moment required by the applicable strength load combination specified in AASHTO Table 3.4.1-1
where:Mr = factored flexural resistanceMcr = cracking momentfr = modulus of rupture
Mcr = Scfr
A Eq. 5.7.3.3.2-1The section modulus is
Sc =Iy
g
b
The modulus of rupture of concrete isA Art. 5.4.2.6
fr = ′0 37. fc
= 0 37 4 5 2. . kipsin
= 0 7849 2. kipsin
40 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Mcr =
( )Iy
fg
tr
=
264 183 631 4
0 78494
2, .
..in
inkipsin
112
ftin
= 550.3 kip-ft
1.2 Mcr = (1.2)(550.3 kip-ft)
= 660.4 kip-ft [controls]
1.33 Mu = (1.33)(2631.9 kip-ft)
= 3500.4 kip-ft
Mr = ΦMn = 0.9(3488.9 kip-ft) = 3140.0 kip-ft > 1.2 Mcr = 660.4 kip-ft [OK]
Design for shear.
The effective shear depth, dv, taken as the distance between the resultants of the tensile and compressive forces due to flexure
A Art. 5.8.2.9
dv = − = − =d a in in ins 238 5 2 45
237 3. . .
where:ds = distance from the extreme compression fiber to the centroid of tensile
reinforcement
The effective shear depth, dv, need not be less than the greater of:A Art. 5.8.2.9
• 0.9 de = 0.9(38.5 in) = 34.6 in < 37.3 in• 0.72 h = 0.72(44 in) = 31.7 in < 37.3 in
The critical section for shear is taken as dv from the internal face of sup-port. The distance from the center of bearing support, x, is calculated as
A Art. 5.8.3.2
x = + =
=37 3 5 7 43 112
3 58. . .in in in ftin
ft
41Design Examples
© 2008 Taylor & Francis Group, LLC
See Figure 2.12.
Find the unfactored shear forces and concurrent moments due to dead load.
Note that the exterior girder controls for moment.The shear at the critical section for shear due to the DC load [interior girder
controls] is
wDC = 1.98 kips/ft
VDC =( )
−
1 98 50
21 98
..
kipsft
ft kipsft
33 58. ft( )
VDC = 42.2 kips
The concurrent moment at the critical section for shear for the interior girder due to the DC load is
MDC =
( ) −50
21 98 3 58 1 98ft kips
ftft ki. . . pps
ftft
( )
3 58 12
2.
= 189.9 kip-ft
The shear at the critical section for shear due to the DW load is
Support
x = 3.58 ftCL
Bearing
dv = 37.3 in5.7 in
Critical section forshear
FIGURE 2.12Critical shear section at support.
42 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
wDW = 0.30 kips/ft
VDW =
( )
−
0 30 50
20 30 3
.. .
kipsft
ftkips
ft558 ft( )
VDW = 6.43 kips
The concurrent moment at the critical section for shear due to the DW load is
MDW =( )
( ) −
50 0 30
23 58 0 30
ft kipsft ft kips
ft
.. .
( )
3 58 12
2. ft
MDW = 24.93 kip-ft
Find the shear at the critical section for shear for the interior girder due to the design truck (HS-20).
MBE-2 App Tbl. H6B*
Let x = L – x; multiply by 2 for two wheel lines.
V =−( )36 9 33
2 kips x
L.
( )
Vtr, HS-20 =− −( )72 9 33 kips L x
L.
=− −( )72 50 3 58 9 33
50 kips ft ft
ft. .
= 53.41 kips per lane
VTL = Vtr(DFV)(1 + IM)
= (53.71 kips)(0.95)(1.33)
= 67.48 kips per beam
* MBE-2 refers to the Manual for Bridge Evaluation (second edition), 2011, American Association of State Highway and Transportation Officials (AASHTO)
43Design Examples
© 2008 Taylor & Francis Group, LLC
Find the concurrent maximum moment per lane interior girder at the criti-cal section for shear due to the design truck (HS-20).
MBE-2 App. Tbl. J6B
Let x = L – x; multiply by 2 for two wheel lines.
M =−( ) −( ) ( )36 9 33
2 kips L x x
L.
Mtr, HS-20 = ( ) − −( )72 9 33 kips x L xL
.
= ( ) − −( )72 3 58 50 3 58 9 3350
kips . . .ft ft ftft
= 191.2 ft-kips per lane
MTL = Mtr(DFM)(1 + IM)
= (191.2 ft-kips)(0.822)(1.33)
= 209.03 ft-kips per beam
Find the shear at the critical section for interior girder due to lane load. See Figure 2.13.
Vln =−( )0 32 2. L x
L
=−( )0 32 50 3 58
50
2. .ft ftft
= 13.79 kips per lane
L = 50 ft
x =3.58 ft
L – x
0.64 kips/ft
FIGURE 2.13Lane load position for maximum shear at critical shear section.
44 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
VLN = Vln(DFV) = (13.79 kips)(0.95)
= 13.1 kips per beam
Find the concurrent moment at the critical section for shear for interior girder due to lane load, where
R =−( )0 32 2. L x
L
Mln = Rx
=−( ) ( )0 32 2. L x xL
=−( ) ( )0 32 50 3 58 3 58
50
2. . .ft ft ftft
= 49.37 ft-kips per beam
MLN = Mln(DFM) = (49.37 ft-kips)(0.822)
= 40.58 ft-kips per beam
The factored shear at the critical section is,A Tbls. 3.4.1-1; 3.4.1-2
Vu = 1.25 VDC + 1.5 VDW + 1.75(VTL + VLN)
= 1.25(42.2 kips) + 1.5(6.43 kips) + 1.75(67.48 kips + 13.1 kips)
Vu = 203.4 kips
The factored moment at the critical section is
Mu = 1.25 MDC + 1.5 MDW + 1.75(MTL + MLN)
= 1.25(189.9 kip-ft) + 1.5(24.93 kip-ft) + 1.75(209.03 kip-ft + 40.58 kip-ft)
Mu = 711.59 kip-ft
Find the shear stress on the concrete.A Eq. 5.8.2.9-1
45Design Examples
© 2008 Taylor & Francis Group, LLC
The effective web width, bv, is 18 in.The resistance factor for shear specified for reinforced concrete is 0.9.
A Art. 5.5.4.2
The distance from extreme compression fiber to the centroid of tensile reinforcement, ds, is 38.5 in (ds is equivalent to de).
A Art. 5.8.2.9
NOT E: de is also the distance from the centerline of the exterior web of the exterior beam to the interior edge of curb or traffic barrier (Art. 4.6.2.2.1).
The factored shear stress on the concrete at dv = 37.3 in is,A Eqs. 5.8.2.9-1, 5.5.4.2.1
vu = = ( )( )( ) =Vb d
kipsin in
ku
v vΦ203 4
0 9 18 37 30 34.
. .. iips
in2
Find the tensile strain in the transverse reinforcement for sections where the transverse reinforcement is found using Eq. 5.8.3.4.2-1 and has the fol-lowing characteristics.
A Art. 5.8.3.4.2θ = 30°
cot θ = 1.732
Es = 29,000 kips/in2
As = 18.75 in2
dv = 37.3 in
εθ
x
u
vu
s s
Md
V
E A=
+ 0 5
2
. cot
A App. B5 Eq. B5.2-1
=
+711 5937 3
12 0 5 2..
.ft-kipsin
inft
003 4 1 732
2 29000 18 752
. .
.
kips
kipsin
i
( )( )
nn2( )
= 0.00037 < 0.001 [OK]
46 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Transverse reinforcement shall be provided where:
A Eq. 5.8.2.4-1; Arts. 5.8.3.4.1
Vu ≥ 0.5 ΦVc
β = 2 and θ = 45° may be used.
The nominal shear resistance, Vn, shall be as the lesser of:
A Art. 5.8.3.3; Eq. 5.8.3.3-1; Eq. 5.8.3.3-2
Vn = Vc +Vs
Vn = 0.25 f′c bvdv
The nominal shear resistance by the concrete is
V f b dc c v v= ′0 0316. β
= ( ) ( )( )0 0316 2 4 5 18 37 32. . .kipsin
in in
= 90.1 kips
Transverse reinforcement shall be provided if Vu ≥ 0.5 ΦVc
A Art. 5.8.2.4
0.5 ΦVc = (0.5)(0.9)(90.1 kips)
= 40.5 kips < Vu = 203.4 kips
Therefore, the transverse reinforcement is provided at the critical section (x = 3.58 ft) for shear.
The nominal shear resistance of a transversely reinforced section is Vn = Vc + Vs. Vs is the shear resistance by reinforcement.
A Eq. 5.8.3.3-1
The nominal shear resistance of the section is
A Eq. 5.8.3.3-2
Vn = ′ = ( )
( )0 25 0 25 4 5 18 372. . .f b d kips
ininc v v ..3 in( )
= 755.3 kips
47Design Examples
© 2008 Taylor & Francis Group, LLC
The factored shear resistance, Vr, isA Eq. 5.8.2.1-2
Vr = ΦVn = (0.9)(755.3 kips)
= 679.8 kips > Vu = 203.4 kips [OK]
The factored shear force does not exceed the maximum factored shear resistance, therefore, the section size is good for shear.
Find the transverse reinforcement requirements.
The maximum shear resistance provided by shear reinforcement isA Art. 5.8.3.3
V V V f b d f b ds n c c v v c v v= − = ′ − ′0 25 0 0316. . β
= 755.3 kips – 90.1 kips
= 665.2 kips
Shear required by shear reinforcement at the critical section is determined by letting the nominal shear resistance, Vn, equal to the factored shear forces, Vu, divided by Φ.
V V V V Vs required n cu
c. = − = −Φ
= −203 40 9
90 1..
.kips kips
= 136.0 kips < 665.2 kips [OK]
θ = angle of inclination of diagonal compressive stress = 45°α = angle of inclination of transverse reinforcement to longitudinal
axis = 90°A Comm. 5.8.3.3, A Eq. 5.8.3.3-4
VA f d
ssv y v=
+(cot cot )sinθ α α
=° + °( ) °( ) =
+A f ds
A f dv y v v y vcot cot sin .45 90 90 1 0 0(( )( )1 0.s
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Try two legs of no. 4 bar stirrups at the critical section, Av = 0.4 in2.
s =A f d
Vv y v
s
=( )
( )0 4 60 37 3
136 0
22. .
.
in kipsin
in
kips
= 6.58 in
Use s = 6.5 in
Minimum transverse reinforcement (where bv is the width of the web) shall be
A Eq. 5.8.2.5-1
Av, min ≥ ′0 0316. f b sfcv
y
Av, min = ( )( )
0 0316 4 5
18 6 5
602
2
. ..kips
inin in
kipsin
= 0.13 in2 < 0.4 in2 at 6.5 in spacing provided [OK]
At the critical section (x = 3.58 ft) from the bearing of the left support, the transverse reinforcement will be two no. 4 bar stirrups at 6.5 in spacing.
Determine the maximum spacing for transverse reinforcement required.A Art. 5.8.2.7
vu is the average factored shear stress in the concrete.A Eq. 5.8.2.9-1
vu = Vb d
u
v vΦ
= ( )( )( )203 4
0 9 18 37 3.
. .kips
in in
= 0 337 2. kipsin
49Design Examples
© 2008 Taylor & Francis Group, LLC
If vu < 0.125 f′c smax = 0.8 dv < 24 in
A Eq. 5.8.2.7-1
vu = <
=0 337 0 125 4 5 0 5632 2. . . .kipsin
kipsin
kippsin2
Thus, smax = 0.8(37.3 in) = 29.8 in ≥ 24 insprovided = 6.5 in ≤ 24 in [OK]
Check tensile capacity of longitudinal reinforcement.
A Art. 5.8.3.5
Shear causes tension in the longitudinal reinforcement in addition to shear caused by the moment.
A Art. 5.8.3.5
At the critical section for shear, x is 3.58 ft.
Mu factored moment 711.59 ft-kips
Vu factored shear 203.4 kips
Φf, Φv resistance factors for moment and shear 0.90
A Art. 5.5.4.2.1
For two no. 4 stirrups at 6.5 in spacing and cot 45° = 1.0, the shear resistance provided by shear reinforcement is,
A Eq. 5.8.3.3-4
Vs = =( )
( )A f d
s
in kipsin
inv y v
0 4 60 37 3
6 5
22. .
. iin
= 137.7 kips
The required tensile capacity of the reinforcement on the flexural tensile side shall satisfy:
A Eq. 5.8.3.5-1
Asfy = + −
Md
V Vu
f v
u
vsϕ ϕ
θ0 5. cot
= ( )( )
+711 590 9 37 3
12 203.. .
ft-kipsin
inft
...
. . .40 9
0 5 137 7 1 0kips kips− ( )
( )
= 411.5 kips
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The available tension capacity of longitudinal reinforcement is
Asfy = ( )
18 75 6022. in kips
in = 1123.2 kips > 411.5 kips [OK]
Step 2: Check the Fatigue Limit StateA Art. 5.5.3
The fatigue strength of the bridge is related to the range of live load stress and the number of stress cycles under service load conditions. The fatigue load combination specified in AASHTO Table 3.4.1-1 is used to determine the allowable fatigue stress range, ff (= constant-amplitude fatigue threshold, (ΔF)TH) (AASHTO Art. 5.5.3.1). The minimum live-load stress level, fmin, is determined by combining the fatigue load with the permanent loads. fmin will be positive if it is in tension.
The factored load, Q, is calculated using the load factors given in AASHTO Tables 3.4.1-1 and 3.4.1-2.
For a simple span bridge with no prestressing, there are no compres-sive stresses in the bottom of the beam under typical dead load conditions. Therefore, fatigue must be considered.
A Art. 5.5.3.1
Fatigue Limit State II (related to finite load-induced fatigue life) Q = 0.75(LL + IM), where LL is the vehicular live load and IM is the dynamic load allowance.
A Tbl. 3.4.1-1
The following information will be used to determine the fatigue load.One design truck that has a constant spacing of 30 ft between 32 kip axles.
A Art. 3.6.1.4.1
Dynamic load allowance, IM, is 15%.A Tbl. 3.6.2.1-1
A distribution factor, DFM, for one traffic lane shall be used.A Art. 3.6.1.4.3b
The multiple presence factor of 1.2 shall be removed.A Comm. 3.6.1.1.2
The allowable fatigue stress range, ff (= (ΔF)TH), shall be equal to (24 – 0.33 fmin).
A Eq. 5.5.3.2-1
51Design Examples
© 2008 Taylor & Francis Group, LLC
ff Allowable fatigue stress range, kips/in2
fmin Minimum live-load stress resulting from the fatigue load com-bined with the permanent loads; positive if tension, kips/in2
See Figure 2.14.The moment per lane due to fatigue load, Mf, is 445.6 ft-kips. For one traffic
lane loaded,
DFMsi = 0.629 lanes for interior girders
DFMse = 0.87 lanes for exterior girders
The multiple presence factor of 1.2 must be removed. Therefore, the distri-bution factor for fatigue load using DFMse is,
A Art. 3.6.1.1.2
DFMfatigue = =0 871 2
0 725..
. lanes
Find the unfactored fatigue load moment per beam.The moment due to fatigue load per beam, Mfatigue, is
Mfatigue = Mf(DFM)(1 + IM)
= (445.6 kip-ft)(0.725)(1 + 0.15)
= 371.5 kip-ft
14 ft
25 ft
Mf = 445.6 kips/ft
25 ft
–x = 2' – 9½"11' – 2½"
12' – 4¾"
R = 21.12 kips
Mf = Moment per lanedue to fatigue load
R = 18.9 kips
23' – 7¼"Span
CL
30 ft8 kips
32kips
C.G. oftwo truck axles on span
32 kips
Centroid of two axles onthe span from32 kips,
x = (8 kip)(14 ft)(32 kips + 8 kips)
–
x = 2.8 ft = 2 ft – 9½–
Fatigue truck loading
FIGURE 2.14Fatigue truck loading and maximum moment at 32 kips position per lane due to fatigue loading.
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Fatigue Investigations.
It is noted that provisions used for Fatigue I are conservative for Fatigue II load.
A Art. 5.5.3.1
For fatigue investigations, the section properties shall be based on cracked sections when the sum of the stresses, due to unfactored permanent loads and Fatigue I load combination, is tensile and exceeds 0.095√f′c. Referring to the unfactored exterior girder moments previously found,
MDC + MDW + Mfatigue = 756.3 kip-ft + 84.4 kip-ft + 371.5 kip-ft
= 1212.2 kip-ft [for exterior girders]
Find the tensile stress at the bottom fiber of gross section of interior girders using the stresses for exterior girders to be conservative.
A Art. 5.5.3.1
Sg =Iy
g
t
= 264 183 631 4
4, ..
inin
= 8,413.5 in3
The tensile stress at the bottom fiber ft is
ft = MSg
=
1212 2 121
8413 5 3
.
.
k t inft
in
ip-f
= 1.73 kips/in2
0 095 0 095 4 5 2. . .′ =f kipsinc
= 0.20 kips/in2 < ft = 1.73 kip/in2
A Art. 5.5.3.1
53Design Examples
© 2008 Taylor & Francis Group, LLC
Therefore, cracked section analysis should be used for fatigue investigation.The modulus of elasticity for concrete with wc = 0.15 kips/in2 is
Ec = ( ) ′33 000 1 5, .w fc c
A Eq. 5.4.2.4-1
=
33 000 0 15 4 53
1 5
2, . ..kips
ftkipsin
Ec = 4,066.8 kips/in2
The modulus of elasticity for steel, Es, is 29,000 kips/in2
The modular ratio between steel and concrete is
n = EE
s
c
= 7.13, use n = 7
Determine the transformed area.
nAs = 7(18.75 in2) = 131.25 in2
Find the factored fatigue moment per beam.
The factored load for Fatigue II load combination, Q, is
A Table 3.4.1-1
Q = 0.75(LL + IM)
MF, fatigue = 0.75(Mfatigue) = 0.75(371.5 kip-ft) = 278.6 kip-ft
See Figure 2.15 for the cracked section analysis.
Find the distance, x, from the top fiber to the neutral axis. Taking the moment of areas about the neutral axis,
1202
131 25 38 52in x x in in x( )( )
= ( ) −( ). .
x = 8.15 in
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INA = +13
3 2bx Ay
= ( )( ) + ( ) −13
120 8 15 131 25 38 5 8 153 2in in in in i. . . . nn( )2
= 142,551.0 in4
Find the stress in the reinforcement.
The stress in the reinforcement due to the factored fatigue live load fs, is,
fs = ( )( )
n
M yI
=
( )nM
IyF fatigue
NA
= ( )( )
7
278 6 121
142 551 4
.
,
ki inft
in
p-ft
−( )38 5 8 15. .in in
fs = 4.98 kips/in2
18 in nAs = 131.25 in2
5.5 in
9 in
ds =38.5 in
120 in
xNA
y = 38.5 in – 8.15 in = 30.35 in
FIGURE 2.15Cracked section determination of T-beam.
55Design Examples
© 2008 Taylor & Francis Group, LLC
Find the fatigue stress range.
The permissible stress range in the reinforcing bars resulting from the fatigue load combination, ff, must not exceed the stress in the reinforcement, fs.
Using the previously calculated unfactored exterior girder moments due to the dead loads, the total moment due to the dead load is
Mdead load = MDC + MDW = 756.3 ft-kips + 84.4 ft-kips
= 840.7 ft-kips
The minimum live load stress resulting from the fatigue load is
fmin = fs, dead load = ( )
n M yI
deadload
= ( ) ( )
−7 840 7 12 38 5 8 15. . .f s ftin
in int-kip (( )
142551 4in
= 15.0 kips/in2
The allowable fatigue stress range ff, is,
A Eq. 5.5.3.2-1
ff = 24 – 0.33 fmin
= 24 – (0.33)(15.0 kips/in2)
= 19.05 kips/in2
fs = stress in the reinforcement due to the factored fatigue live load
= 4.98 kips/in2 < ff = 19.05 kips/in2 [OK]
The stress in the reinforcement due to the fatigue live load is less than the allowable fatigue stress range, so it is ok.
Step 3: Check the Service I Limit State
The factored load, Q, is calculated using the load factors given in AASHTO Table 3.4.1-1.
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The load combination relating to the normal operational use of the bridge with all loads taken at their nominal values (with no impact allowance, no load factors, and so on), is similar to that for the allowable stress methods. This combination is for deflection control and crack width control in rein-forced concrete.
Q = 1.0(DC) + 1.0(DW) + 1.0(TL + LN)
A Tbl. 3.4.1-1
Although deflection and depth limitations are optional in AASHTO, bridges should be designed to avoid undesirable structural or psychological effects due to deformations.
A Art. 2.5.2.6.1
Criteria for deflection in this section shall be considered optional. However, in the absence of other criteria, the following limits may be considered.
A Art. 2.5.2.6.2
For steel, aluminum, concrete vehicular bridges,
vehicular load general span/800
For wood vehicular bridges,
vehicular and pedestrian loads span/425
vehicular load on wood planks and panels
(extreme relative deflection between adjacent edges) 0.10 in.
Using the previously calculated exterior girder moments due to the dead and live loads, the total service load moment is
Mservice = 1.0 MDC +1.0 MDW + 1.0(MTL + MLN)
= (1.0)(756.3 ft-kips) + (1.0)(84.4 ft-kips)
+ (1.0)(717.4 ft-kips + 174.0 ft-kips)
= 1732.1 ft-kips
Find the control of cracking by distribution of reinforcement.
The maximum spacing of tension reinforcement applies if the tension in the cross section exceeds 80% of the modulus of rupture, fr, at the Service I Limit State.
A Art. 5.7.3.4
57Design Examples
© 2008 Taylor & Francis Group, LLC
The modulus of rupture isA Art. 5.4.2.6
fr = ′0 24. fc
= 0 24 4 5 2. . kipsin
= 0.51 kips/in2
The section modulus at the bottom fiber for the gross cross section (where yt is equivalent to yb) is
Sg =Iy
g
t
where:yt = distance from the neutral axis to the extreme tension fiber = y + 5.5 in = 35.85 in
= 264 183 635 85
4, ..
inin
= 7369.1 in3
The tensile stress at the bottom fiber is
ft = =( )
M
S
t inftservice
g
1732 1 121
7369
. kip-f
..1 3in
= 2.82 kips/in2 > 0.8 fr = 0.8(0.51 kips/in2) = 0.41 kips/in2
Thus, flexural cracking is controlled by limiting the spacing, s, in the ten-sion reinforcement. Therefore, the maximum spacing, s, of the tension rein-forcement should satisfy the following.
A Eq. 5.7.3.4-1
s ≤ −700 2γβ
e
s sscf
d
The exposure factor for a class 2 exposure condition for concrete, γe, is 0.75. The overall thickness, h, is 44 in. The concrete cover measured from extreme tension fiber to the center of the flexural reinforcement located clos-est thereto is
A Art. 5.7.3.4
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dc = 1.5 in + 0.5 in + 1.41 in/2
= 2.7 in (see Figure 2.11)
The modular ratio between the steel and concrete, n, was previously cal-culated as 7. The tensile stress in steel reinforcement at the Service I Limit State is
A Art. 5.7.3.4
fss =
n M
Sservice
g,
where:
SIygg= = =264 183 6
30 358704 6, .
.. in
in in
43
=
( )
7
1732 1 121
8 413 5 3
.
, .
k t inft
in
ip-f
= 16.7 kips/in2
The ratio of the flexural strain at extreme tension face to the strain at the centroid of reinforcement layer nearest to the tension face is
A Art. 5.7.3.4
βs = +−( )
1
0 7dh d
c
c.
= +−( )
1 2 7
0 7 44 2 7.
. .in
in in
= 1.09
Therefore, the maximum spacing, s, of the tension reinforcement shall sat-isfy the following.
s ≤ −700 2γβ
e
s sscf
d
≤ ( )( )
− ( )700 0 75
1 09 16 72 2 7
2
.
. ..
kipsin
in
≤ 24.3 in
59Design Examples
© 2008 Taylor & Francis Group, LLC
Provided bar spacing, s, is 4.19 in, which is less than the maximum spacing allowed so it is ok (see Figure 2.11).
Step 4: Design the Deck Slab
In AASHTO, concrete decks can be designed either by the empirical method or the traditional method.
A Art. 9.7
The empirical design method may be used for concrete deck slabs sup-ported by longitudinal components if the following conditions are satisfied.
A Arts. 9.7.2, 9.7.2.4
a. The supporting components are made of steel and/or concrete.
b. The deck is fully cast-in-place and water cured.
c. The deck is of uniform depth, except for haunches at girder flanges and other local thickening.
d. The ratio of effective length to design depth does not exceed 18 and is not less than 6.0.
The effective length is face-to-face of the beams monolithic with slab; therefore,
ratio effective length
design depthft ft= = −10 1 5.
99 112
in ftin( )
= 11.33 < 18 > 6 [OK]
e. Core depth of the slab (out-to-out of reinforcement) is not less than 4 in.
Assuming clear cover of 2.5 in for top bars and 1.5 in for bottom bars,
9 in – 2.5 in – 1.5 in = 5 in > 4 in [OK]
f. The effective length does not exceed 13.5 ft.
8.5 ft < 13.5 ft [OK]
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g. The minimum depth of slab is not less than 7 in.
9 in > 7 in [OK]
h. There is an overhang beyond the centerline of the outside girder of at least 5 times the depth of slab. This condition is satisfied if the overhang is at least 3 times the depth of slab and a structurally con-tinuous concrete barrier is made composite with the overhang.
Overhang = (4.0 ft)(12 in/ft).Check requirements.
(5)(9 in) = 45 in < 48 in [OK]
(3)(9 in) = 27 in < 48 in [OK]
i. The specified 28-day strength of the deck concrete, f′c , is not less than 4 kips/in2.
f′c = 4.5 kips/in2 > 4 kips/in2 [OK].
j. The deck is made composite with the supporting structural elements.
Extending the beam stem stirrups into deck will satisfy this requirement [OK].
The empirical method may be used because all of the above conditions are met.
Four layers of isotropic reinforcement shall be provided. The minimum amount of reinforcement in each direction shall be 0.27 in2/ft for bottom steel, and 0.18 in2/ft for top steel. Spacing of steel shall not exceed 18 in.
A Art. 9.7.2.5
Bottom reinforcement: no. 5 bars at 12 in; As = 0.31 in2 > 0.18 in2 [OK].Top reinforcement: no. 5 bars at 18 in; As = 0.20 in2 > 0.18 in2 [OK].The outermost layers shall be placed in the direction of the slab length.
Alternatively, the traditional design method may be used.
A Art. 9.7.3
If the conditions for the empirical design are not met, or the designer chooses not to use the empirical method, the LRFD method allows the use of the traditional design method.
61Design Examples
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Concrete slab shall have four layers of reinforcement, two in each direction and clear cover shall comply with the following AASHTO Art. 9.7.1.1 provisions.
Top bars have 2.5 in and bottom bars have 1.5 in clear covers.The approximate method of analysis for decks divides the deck into
strips perpendicular to the support elements. The width of the strip is calculated according to AASHTO Art. 4.6.2.1.3.
A Art. 4.6.2.1
The width of the primary strip is calculated for a cast-in-place concrete deck using the following calculations from AASHTO Table 4.6.2.1.3-1.
Overhang: 45 + 10 X+M: 26 + 6.6 S–M: 48 + 3.0 S
The spacing of supporting elements, S, is measured in feet. The distance from load to point of support, X, is measured in feet.
A deck slab may be considered as a one-way slab system. The strip model of the slab consists of the continuous beam, and the design truck is posi-tioned transverse for the most critical actions with a pair of 16 kip axles spaced 6 ft apart.
Design Example 2: Load Rating of Reinforced Concrete T-Beam by the Load and Resistance Factor Rating (LRFR) Method
Problem Statement
Use the Manual for Bridge Evaluation (MBE-2), Second Edition 2011, Section 6: Load Rating, Part A Load and Resistance Factor Rating (LRFR). Determine the load rating of the reinforced-concrete T-beam bridge interior beam using the bridge data given for the Limit State Combination Strength I. Also refer to FHWA July 2009 Bridge Inspection System, Condition and Appraisal rating guidelines.
L span length 50 fts beam spacing 10 ftf′c concrete strength 4.5 kips/in2
fy specified minimum yield strength of steel 60 kips/in2
DW future wearing surface load 0.03 kips/ft2
ADTT average daily truck traffic in one direction 1900 skew 0°Year Built: 1960
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Condition: Good condition with some minor problemsNBIS Item 59, Condition Good (Code 7)
See Figures 2.16 and 2.17.
Solution
Step 1: Analysis of Dead Load Components—Interior Beam
1.A Components and Attachments – DC
DCT-beam – Structural = 1.78 kips/ft
DCCurbs & Parapet = 0.202 kips/ft
Total DC per beam = DCT-beam – Structural + DCCurbs & Parapet
= 1.78 kips/ft + 0.202 kips/ft
Total DC per beam = 1.98 kips/ft
24 ft24 ft
10 ft 10 ft 10 ft 4 ft4 ft
2 ft 8 in
1 ft 9 in1 ft 9 in
ts = 9 inCurb and parapetRoadway = 44 ft 6 in
S = 10 ft
FIGURE 2.16T-beam bridge cross section.
bv = 18"
35"
9"
12 # 11 bars(As = 18.75 in2)
120 in
FIGURE 2.17T-beam section.
63Design Examples
© 2008 Taylor & Francis Group, LLC
M w Lkips
ftft
kDCDC= =
( )
=2
2
8
1 98 50
8618 75
.. iip-ft
V w Lkips
ftft
kipsDCDC= =
( )
=2
1 98 50
249 5
..
1.B Wearing Surface – DW
w kipsft
ftbeam
kipsDW =
=0 03 10 0 32. .fft
M w Lkips
ftft
kipDWDW= =
( )
=2
2
8
0 3 50
893 75
.. --ft
V w Lkips
ftft
kipsDWDW= =
( )
=2
0 3 50
27 5
..
Step 2: Analysis of Live Load Components—Interior Beam
2.A Live Load Distribution Factor
AASHTO Cross-Section Type (e)A Tbl 4.6.2.2.1-1
For preliminary design,A Tbl. 4.6.2.2.1-2
KLtg
s121 053
0 1
=
.
.
2.A.1 Distribution Factor for Moment in Interior Beam, DFM
A Tbl 4.6.2.2b-1 or Appendix A
i) One lane loaded, DFMsi
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DFM ft ftftsi = +
0 06 10
141050
0 4 0 3
.. .
11 05.
DFMsi = 0.626 lanes
ii) Two or more lanes loaded, DFMmi
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
DFM ft ftftmi = +
0 075 10
9 51050
0 6 0
..
. ..
.2
1 05
DFMmi = 0.860 lanes [controls]
2.A.2 Distribution Factor for Shear in Interior Beam, DFV
AASHTO [Tbl 4.6.2.2.3a-1] or Appendix C
i) One lane loaded, DFVsi
DFV Ssi = +0 36
25.
DFV ftsi = +0 36 10
25.
DFVsi = 0.76 lanes
ii) Two or more lanes loaded, DFVmi
DFV S Smi = + −
0 212 35
2 0
..
DFV ft ftmi = + −
0 2 1012
1035
2 0
..
DFVmi = 0.95 lanes [controls]
65Design Examples
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2.B Maximum Live Load Effects
2.B.1 Maximum Live Load (HL-93) Moment at Midspan (see Design Example 1)
Design Truck Moment = Mtr = 620.0 kip-ft
Design Lane Moment = Mlane = 200.0 kip-ft
Design Tandem Axles Moment = 575.0 kip-ft
IM = 33%A Tbl. 3.6.2.1-1
MLL+IM = Mtr(1 + IM) + Mlane
= 620.0 kip-ft(1 + 0.33) +200.0 kip-ft
MLL+IM = 1024.6 kip-ft
2.B.2 Distributed live load moment per beam
MLL+IM per beam = (MLL+IM)(DFM)
= (1024.6 kip-ft)(0.860)
MLL+IM per beam = 881.16 kip-ft
Step 3: Determine the Nominal Flexural Resistance, Mn
3.A Effective Flange Width of T-Beam, be
Effective flange width is the width of the deck over girders.A Comm. 4.6.2.6.1
Average spacing of beams = (10 ft)(12 in/1 ft) = 120.0 in
Use be = 120.0 in
3.B Determination of Nominal Flexural Resistance in Bending, Mn, is
A Eq. 5.7.3.2.2-1
M A f d an s y s= −
2
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As for 12 - #11 bars = 18.75 in2
be = 120 in
β1 0 85 40001000
0 05= − ′ −
( ). .fc
A Art. 5.7.2.2
β1
20 85
4500 4000
10000 05= −
−
( ). .
lbfin
β1 = 0.825
cA f
f bs y
c=
′0 85 1. β
A Eq. 5.7.3.1.1-4
cin kips
inkipsin
=( )
18 75 60
0 85 4 5
22
2
.
. . ( )( )0 825 120. in
c = 2.97 in < ts = 9 in
a = cβ
A Art. 5.7.2.2
a = (2.97 in)(0.825)
a = 2.45 in
ds distance from extreme compression fiber to
the centroid of the tensile reinforcement 38.5 in
See Figure 2.18.
67Design Examples
© 2008 Taylor & Francis Group, LLC
M A f d an s y s= −
2
M in kipsin
in inn = ( )
−18 75 60 38 5 2 452
22. . .
Mn = 41934 kip-in(1 ft/12 in)
Mn = 3494.5 kip-ft
Step 4: Determine the Shear at the Critical Section near Support
A Art 5.8.3.2
Vertical Stirrups: #4 bars at 6.5 in spacing
A A x in inv bar= =
=2 2
448
0 404
22
# .π
The location of the critical section for shear shall be taken as dv from the bearing face of the support.
The effective shear depth, dv, can be determined as
A Eq. C.5.8.2.9-1; 5.8.2.9
ds = Distance from extreme compression fiber to the centroid of tensile reinforcement = 44 in – 5.5 in
ds = 38.5 in
18 in 5.5 in
+ C.G.35 in 44 in
InteriorT-beam section
9 in3.6 in
yt = 31.4 in
be = 120 in
FIGURE 2.18Interior T-beam section for determination of flexural resistance.
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d MA fv
n
s y=
dki t in
ft
in kv =( )
( )
3494 5 121
18 74 602
.
.
p-f
iipsin2
dv = 37.3 in [controls]
But dv need not be taken less than to be less than the greater of 0.9 de or 0.72 h:
A Art. 5.8.2.9
dv = 0.9 de(= ds)
= 0.9(38.5 in)
dv = 34.7 in
or
dv = 0.72 h
= 0.72(44 in)
dv = 31.7 in
Use dv = 37.3 in.
Check: dv = distance between the resultants of the tensile and compressive force due to flexure,
d d a in inv s= − = −
238 5 2 45
2. .
= 37.3 in (check OK)
See Figure 2.19.The critical section for shear from face of bearing support, dv, is 37.3 in.
Calculate shear at x = 37.3 in +5.7 in = 43 in from the center of bearing.
Shear at x = (43 in) (1 ft/12 in) = 3.58 ft from the support end.
69Design Examples
© 2008 Taylor & Francis Group, LLC
Dead Load wDC = 1.98 kips/ft
wDW = 0.3 kips/ft
Maximum live load shear at critical section on a simple spanMBE-2 App. Tbl. H6B
Live load shear at critical section due to design truck load, HS-20,
Let x = L – x; multiply by 2 for two wheel lines.
HS-20: Vkips x
Lx=
−( )36 9 332
. and x = (L-x)
Vkips L x
L=
−( ) − 72 9 33.
Vtr = HS-20 at x = 3.58 ft:
Vkips ft ft
fttr HS= =−( ) −
-2072 50 3 58 9 33
50. .
Vtr = HS-20 = 53.41 kips per lane
VTL = (Vtr)(DFV)(1 + IM)
= (53.41 kips per lane)(0.95)(1.33)
VTL = 67.5 kips per beam
11.4 in
A
Span = L = 50 ft
dv = 37.3 in
x = (5.7 in + 37.3 in) (1 ft/12 in) = 3.58 ft
5.7 in
Bearing cFIGURE 2.19Critical section for shear at support.
70 Simplified LRFD Bridge Design
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Live load shear at critical shear section due to lane load (see Figure 2.20),
R
kipsft
ft ft
A =
( )
0 64 46 42 46 422
5
. . .
00 ft
RA = 13.79 kips
Vln@x = 3.58 ft = 13.79 kips
VLN = Vln(DFV) = (13.79 kips)(0.95) = 13.1 kips per beam
VLL+IM = total live load shear = 67.5 kips + 13.1 kips = 80.6 kips
Nominal Shear Resistance, Vn, shall be the lesser of:A Art. 5.8.3.3
Vn = Vc + Vs
A Eq. 5.8.3.3-1, A Eq. 5.8.3.3-2
V f b dn c v v= ′0 25.
V f b dc c v v= ′0 0316. β
VA f d
ssv y v=
cotθ
α = 90°A Comm. 5.8.3.3-1
For simplified procedure,A Art. 5.8.3.4.1
L = 50 ft
RA = 13.79 kips46.42 ft
A0.69 f/ft
15.91 kipsx = 3.58 ft
x
FIGURE 2.20Live load shear at critical shear section due to lane load.
71Design Examples
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β = 2.0; θ = 45°
V kipsin
in inc = ( ) ( )( )0 0316 2 0 4 5 18 37 32. . . .
Vc = 90.0 kips
Vin kips
inin
s =( )
( ) °0 40 60 37 3 45
6
22. . cot
.55 in
Vs = 137.7 kips
Vn = 90.0 kips + 137.7 kips = 227.7 kips [controls]
V f b dn c v v= ′0 25.
V kipsin
in inn =
( )( )0 25 4 5 18 37 32. . .
Vn = 356.1 kips
Step 5: Summary of Capacities and Demands for Interior Concrete T-Beam
See Table 2.3.
Step 6: Calculation of T-Beam Load Rating
MBE-2 Eq. 6A.4.2.1-1
RFC DC DW P
LL IMDC DW P
LL=
− ( )( ) − ( )( ) ± ( )( )( ) +(
γ γ γγ ))
TABLE 2.3
Dead Loads and Distributed Live Loads Effects Summary for Interior T-Beam
DC DWLL Distribution
FactorDistributed LL+Impact
Nominal Capacity
Moment, kip-ft 618.75 93.75 0.860 881.16 3494.5Shear, kipsa 42.4 6.4 0.95 80.6 227.7
a At critical section (= 3.58 ft from bearing).
72 Simplified LRFD Bridge Design
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For the strength limit states:MBE-2 Eq. 6A.4.2.1-2
C = ΦcΦsΦRn
where:RF = rating factorC = capacityfR = allowable stress specified in the LRFD codeRn = nominal member resistance (as inspected)DC = dead load effect due to structural components and attachmentsDW = dead load effect due to wearing surface and utilitiesP = permanent loads other than dead loadsLL = live load effectIM = dynamic load allowanceγDC = LRFD load factor for structural components and attachmentsγDW = LRFD load factor for wearing surfaces and utilitiesγp = LRFD load factor for permanent loads other than dead loads = 1.0γLL = evaluation live load factorΦc = condition factor
MBE-2 Tbl 6A.4.2.3-1Φc = 1.0 for good conditionΦs = system factor
MBE-2 Tbl 6A.4.2.4-1Φs = 1.0 for all other girder bridgesΦ = LRFD resistance factor
A Art. 5.5.4.2.1Φ = 0.90 for shear and flexure
6.A Strength I Limit State
RFR DC DW
LLc s n DC DW
LL= ( )( )( ) − ( )( ) − ( )( )
( )Φ Φ Φ γ γ
γ ++( )IM
MBE-2 Eq. 6A.4.2-1
Please see Table 2.4.
TABLE 2.4 (MBE-2 Tbl 6A.4.2.2-1)
Load Factors for Load Rating for Reinforced Concrete Bridge
Bridge Type Limit StateDead Load,
γDC
Dead Load, γDW
Design Live Load
Inventory OperatingγLL γLL
Reinforced Concrete Strength I 1.25 1.5 1.75 1.35
73Design Examples
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6.B Load Rating for Inventory Level
Flexure:
RF
ki t
=
( ) ( )( )( ) − ( )1 0 1 0 0 9 3494 5 1 25 61. . . . .p-f 88 75
1 50 93 751 75 88
.
. ..
ki
t
p-ft
kip-f( )
−( )( )( ) 11 16. ki tp-f( )
RF kiki t
= 2231 01542 03
..
p-ftp-f
RF = 1.45
Shear:
RFkips kip
= ( )( )( )( ) − ( )1 0 1 0 0 9 227 0 1 25 42 4. . . . . . ss kipskips( ) − ( )( )
( )( )1 50 6 4
1 75 80 6. .
. .
RF kipskips
= 141 7141 05
..
RF = 1.00
6.C Operating Level Load Rating
Flexure:
RF = ( )
=1 45 1 751 35
1 88. ..
.
Shear:
RF = ( )
=1 00 1 751 35
1 30. ..
.
6.D Summary of Rating Factors for Load and Resistance Factor Rating Method (LRFR)—Interior Beam for Limit State Strength I
Please see Table 2.5.
TABLE 2.5
Rating Factor (RF)
Design Load Rating
Limit State Force Effect Inventory Operating
Strength I Flexure 1.45 1.88Shear 1.00 1.30
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6.E Loading in Tons, RT
MBE-2 Eq. 6A.4.4-1
RT = RF x W
where:RF = rating factor (Table 2.5)RT = rating in tons for trucks used in computing live load effectW = weight in tons of truck used in computing live load effect
Design Example 3: Composite Steel–Concrete Bridge
Situation
BW barrier weight 0.5 kips/ftf′c concrete strength 4 kips/in2
ADTT average daily truck traffic in one direction 2500 design fatigue life 75 yrwFWS future wearing surface load 25 lbf/ft2
de distance from the centerline of the exterior web of exterior beam to the interior edge of curb or traffic barrier 2 ftL span length 40 ft load of stay-in-place metal forms 7 lbf/ft2
S beam spacing 8 ftwc concrete unit weight 150 lbf/ft3
fy specified minimum yield strength of steel 60 kips/in2
ts slab thickness 8 in
Requirements
Design the superstructure of a composite steel–concrete bridge for the load combinations Strength I, Service II, and Fatigue II Limit States using the given design specifications.
SolutionAISC 13th Ed. Tbl. 1-1*
For the bare steel section, W24x76, please see Figure 2.21.
* Refers to the American Institute of Steel Construction.
75Design Examples
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Find the effective flange widthsA Art. 4.6.2.6
The effective flange width for interior beams is equal to the beam spacing (see Figure 2.22)
be,int = (8 ft)(12 in/ft) = 96 in
Beam detailsW24 × 76
2 in haunch
Beam cross section
3 ft 3 in S = 8 ft
de = 2 ft
8 in = ts
S = 8 ft
22 ft1 ft 3 in
4 ft
ts = ?? in
CL
FIGURE 2.21Composite steel–concrete bridge example.
D = 22.54 ind = 23.9 in
9 in
0.44 in
0.68 in
W24 × 76tf = 0.68 in
A = 22.4 in2
Ix = 2100 in4
w = 76 lbf/ft
bf = 9 intw = 0.44 ind = 23.9 in
FIGURE 2.22Steel section.
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The effective flange width for exterior beams is equal to half of the adjacent beam spacing plus the overhang width
be,ext = S/2 + overhang = (96 in)/2 + 39 in = 87 in
Find the dead load moments and shears. See Figure 2.23.Find the noncomposite dead load, DC1, per interior girder.
DC in ftin
in ftinslab = ( )
( )
96 112
8 112
=0 15 0 83. . /kipsft
kips ft
DChaunch = ( )
( )
2 112
9 112
0 15in ftin
in ftin
ki. ppsft
kips ft3 0 02
= . /
Assuming 5% of steel weight for diaphragms, stiffeners, and so on,
DCsteel = + ( )
=0 076 0 05 0 076 0 08. . . .kipsft
kipsft
kipps ft/
DCstay-in-place forms
=
( )
7 162
lbfft
roadway widthgirders
==
( )
0 007 44 162. kips
ftft
= 0.051 kips/ft
The noncomposite dead load per interior girder is
9 in
W24 × 76
Width = 0.44 in d = 23.9 in
2 in
96 in
FIGURE 2.23Composite steel section.
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DC1 = DCslab + DChaunch + DCsteel + DCstay-in-place forms
= 0.8 kips/ft + 0.02 kips/ft + 0.08 kips/ft + 0.051 kips/ft
= 0.951 kips/ft
The shear for the noncomposite dead load is
V wLkips
ftft
kipsDC1 2
0 951 40
219 02= =
( )
=.
.
The moment for the noncomposite dead load is
M wLkips
ftft
fDC1
22
8
0 951 40
8190 2= =
( )
=.
. t--kips
Find the composite dead load, DC2, per interior girder.
DC DC
kipsft
barriers
gibarrier2
0 5 2
6= =
( ).
rrderskips ft= 0 167. /
The shear for the composite dead load per interior girder is
V wLkips
ftft
kipsDC2 2
0 167 40
23 34= =
( )
=.
.
The moment for the composite dead load per interior girder is
M wLkips
ftft
fDC2
22
8
0 167 40
833 4= =
( )
=.
. t-kkips
The shear for the total composite dead load for interior girders is
VDC = VDC1 + VDC2 = 19.02 ft-kips + 3.34 ft-kips = 22.36 kips
The moment for the total composite dead load for interior girders is
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MDC = MDC1 + MDC2 = 190.2 ft-kips + 33.4 ft-kips = 223.6 ft-kips
The future wearing surface dead load, DW, for interior girders is
DW w Lno of beams
lbfft
ftFWS
FWS= =
( )
.
25 44
6
2
bbeamskips ft= 0 183. /
The shear for the wearing surface dead load for interior girders is
V wLkips
ftft
kipsDW = =
( )
=2
0 183 40
23 66
..
The moment for the wearing surface dead load for interior girders is
M wLkips
ftft
fDW = =
( )
=2
2
8
0 183 40
836 6
.. t-kiips
Assume that the dead load for the exterior girders is the same as for inte-rior girders. This is conservative.
Dead load summary
VDC = 22.4 kipsMDC = 223.6 ft-kipsVDW = 3.66 kipsMDW = 36.6 ft-kips
Find the live loads.Find the unfactored moments, unfactored shears, and distribution factors for the live loads.
A Art. 4.6.2.2.1
The modulus of elasticity of concrete where wc = 150 lbf/ft3 (0.15 kips/ft3) isA Eq. 5.4.2.4-1
E w f kipsftc c c= ( ) ′ = ( )
33 000 33 000 0 151 5
3, , ..
1 5
24. kips
in
= 3834 kips/in2
79Design Examples
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The modulus of elasticity for the steel, Es, is 29,000 kips/in2.The modular ratio between steel and concrete is
n EE
kipsin
kipsin
s
c= = =
29 000
38347 56
2
2
,.
Use n = 8.
The distance between the centers of gravity of the deck and beam, eg, is 17.95 in.
AISC 13th Ed. Tbl. 1-1
The moment of inertia for the beam, I, is 2100 in4..The area, A, is 22.4 in2.Using the previously calculated values and the illustration shown, calcu-
late the longitudinal stiffness parameter, Kg. See Figure 2.24.The longitudinal stiffness parameter, Kg, is
A Eq. 4.6.2.2.1-1
Kg = n(I + Aeg2) = (8)(2100 in4 + (22.4 in2)(17.95 in)2)
= 74,539 in4
23.9 in
4 in4 in 8 in
2 inN.A. of deck
11.95 in
11.95 in + 2 in = 13.95 in ag = 13.9 in + 4 in= 17.95 in
c.g. beam
9 in
FIGURE 2.24Composite section for stiffness parameter, Kg.
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© 2008 Taylor & Francis Group, LLC
For cross-section type (a):A Tbl. 4.6.2.2.1-1
S = 8 ft
L = 40 ft
ts = 8 in
The distribution factor for moments for interior girders with one design lane loaded is
A Tbl. 4.6.2.2.2b-1
DFM S SL
KLtsig
s= +
0 06
14 12
0 4 0 3
3.. .
0 1.
= +
0 06 814
840
74 5390 4 0 3
. ,. .ft ft
ftinn
ft in
4
3
0 1
12 40 8( )( )
.
= 0.498 lanes
The distribution factor for moments for interior girders with two or more design lanes loaded is
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
= +
0 075 89 5
840
74 530 6 0 2
..
,. .ft ft
ft99
12 40 8
4
3
0 1
inft in( )( )
.
= 0.655 lanes [controls]
The distribution factor for shear for interior girders with one design lane loaded is
A Tbl. 4.6.2.2.3a-1
DFV S ftsi = +
= +
0 3625
0 36 825
. .
= 0.68 lanes
81Design Examples
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The distribution factor for shear for interior girders with two or more design lanes loaded is
DFV S S ft ftmi = + −
= + −
0 212 35
0 2 812
835
2
. .
2
= 0.814 lanes [controls]
The distribution factor for moments for exterior girders with one design lane loaded can be found using the lever rule.
A Art. 6.1.3.1; A Tbl. 4.6.2.2.2d-1 or Appendix B
The multiple presence factors, m, must be applied where the lever rule is used. Please see Figure 2.25.
A Comm. 3.6.1.1.2; Tbl. 3.6.1.1.2-1
ΣM@hinge = 0
=
( ) +
( ) − ( )P ft P ft R ft
28
22 8
R = 0.625 P = 0.625
A multiple presence factor, m, for one lane loaded is applied to the distri-bution factor for moments of exterior girders.
A Tbl. 3.6.1.1.2-1
8 ft2 H
6 ftde = 2 ft 2 ft
Hinge
P2
R3 ft 3 in
P2
FIGURE 2.25Lever rule for determination of distribution factor for moment in exterior beam, one lane loaded.
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DFMse = (DFMse) (m) = R(m)
DFMse = (0.625)(1.2) = 0.75 lanes [controls]
The distribution factor for moments for exterior girders with two or more design lanes loaded using g, a distribution factor is
A Tbl. 4.6.2.2.2d-1 or Appendix B
The distribution factor for interior girder, gint, is 0.655.The correction factor for distribution, where the distance from the center-
line of the exterior web of the exterior beam to the curb, de, is 2 ft, is
e d fte= + = + =0 779 1
0 77 29 1
0 9898..
..
.
DFMme = g = (e)(gint) = (0.9898)(0.655) = 0.648 lanes [controls]
Or, stated another way,
DFMme = (e)DFMmi = (0.9898)(0.655) = 0.648 lanes
The distribution factor for moments for exterior girders with one design lane loaded is 0.75 (> 0.648).
Use the lever rule to find the distribution factor for shear in exterior gird-ers for one design lane loaded. This is the same as DFMse for one design lane loaded (= 0.75).
A Tbl. 4.6.2.2.3b-1 or Appendix D
DFVse = 0.75
Find the distribution factor for shear for exterior girders with two or more design lanes loaded using g, a distribution factor.
e d fte= + = + =0 610
0 6 210
0 8. . .
g = (e)(gint) = (0.8)(0.814) = 0.6512 lanes
Or, stated another way,
DFVme = e(DFVmi) = (0.8)(0.814) = 0.6512 lanes
Find the unfactored live load effects with dynamic load allowance, IM.
83Design Examples
© 2008 Taylor & Francis Group, LLC
The HL-93 live load is made of either the design truck or the design tan-dem load (whichever is larger) and the design lane load.
A Art. 3.6.1.2
Find the design truck moment due to the live load. See Figure 2.26.
ΣM@B = 0
R ft kips ft kips ft kipA 40 32 34 32 20 8( ) = ( )( ) + ( )( ) + ss ft( )( )6
RA = 44.4 kips
The design truck (HS-20) moment due to live load is
Mtr = (44.4 kips)(20 ft) – (32 kips)(14 ft) = 440 ft-kips
Find the design tandem moment due to the live load. Please see Figure 2.27.
ΣM@B = 0
R kips kips ftftA =
+ ( )
252
25 1640
RA = 22.5 kips
The design tandem moment due to the live load is
Mtandem = (22.5 kips)(20 ft) = 450 ft-kips [controls]
40 ft
20 ft20 ft
6 ft32 kips 32 kips 8 kips
6 ft14 ft 14 ft
R
A B
FIGURE 2.26Load position for moment at midspan for design truck load (HS-20).
84 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Find the design lane moment due to the live load. Please see Figure 2.28.
M wLkips
ftft
f ipln
.= =
( )
=2
2
8
0 64 40
8128 t-k ss
The dynamic load allowance, IM, is 33% (applied to the design truck or the design tandem only, not to the design lane load).
A Art. 3.6.2.1
The total unfactored live load moment is
MLL+IM = Mtandem(1 + IM) + Mln = (450 ft-kips)(1 + 0.33) + 128 ft-kips
= 726.5 ft-kips per lane
40 ft
0.64 kips/ft
FIGURE 2.28Load position for moment at midspan for design lane load.
20 ft20 ftR
A B
25 kips
16 ft4 ft
25 kips
FIGURE 2.27Load position for moment at midspan for design tandem load.
85Design Examples
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Find the shear due to the design truck live load. Please see Figure 2.29.
ΣM@B = 0
R ft kips ft kips ft kipA 40 32 40 32 26 8( ) = ( )( ) + ( )( ) + ss ft( )( )12
RA = 55.2 kips (controls)
The shear due to design truck load, Vtr, is 55.2 kips [controls].
Find the shear due to the design tandem load. Please see Figure 2.30.
ΣM@B = 0
R ft kips ft kips ftA 40 25 40 25 36( ) = ( )( ) + ( )( )
RA = 47.5 kips
40 ft
4 ft25 kips25 kips
R
A B
FIGURE 2.30Load position for shear at support design tandem load.
A B
R40 ft
14 ft 14 ft 12 ft8 kips32 kips32 kips
FIGURE 2.29 Load position for shear at support for design truck load (HS-20).
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The shear due to the design tandem load, Vtandem, is 47.5 kips.
Find the shear due to the design lane load. Please see Figure 2.31.
V V wLkips
ftft
kips= = =
( )
=ln
..
2
0 64 40
212 8
The dynamic load allowance, IM, is 33% (applied to the design truck or the design tandem only, but not to the design lane load).
The total unfactored live load shear is
VLL+IM = Vtr(1 + IM) + Vln = (55.2 kips)(1.33) + 12.8 kips
= 86.22 kips per lane
Please see Table 2.6.
Find the factored moments and shears.
A Tbls. 3.4.1-1; 3.4.1-2
40 ft
0.64 kips/ft
FIGURE 2.31Load position for shear at support for design lane load.
TABLE 2.6
Complete Live Load Effect Summary
Girder Location
No. of Design Lanes
Loaded
Unfactored MLL+IM
(ft-kips per Lane) DFM
Unfactored MLL+IM
(ft-kips per Girder)
Unfactored VLL+IM
(kips per Lane) DFV
Unfactored VLL+IM
(kips per Girder)
Interior 1 726.5 0.498 — 86.22 0.68 —2 726.5 0.655 475.9 86.22 0.814 70.2
Exterior 1 726.5 0.75 544.9 86.22 0.75 64.72 726.5 0.648 — 86.22 0.6512 —
87Design Examples
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Check the Strength I Limit State.
U = 1.25 DC + 1.5 DW + 1.75(LL + IM)
The factored moment for interior girders is
Mu = 1.25 MDC + 1.5 MDW + 1.75 MLL+IM
= (1.25)(223.6 ft-kips) + (1.5)(36.6 ft-kips) + (1.75)(475.9 ft-kips)
= 1167.2 ft-kips
The factored shear for interior girders is
Vu = 1.25 VDC + 1.5 VDW + 1.75 VLL+IM
= (1.25)(22.4 kips) + (1.5)(3.66 kips) + (1.75)(70.2 kips)
= 156.3 kips
The factored moment for exterior girders is
Mu = (1.25)(223.6 ft-kips) + (1.5)(36.6 ft-kips) + (1.75)(544.9 ft-kips)
= 1288 ft-kips
The factored shear for interior girders is
Vu = (1.25)(22.4 kips) + (1.5)(3.66 kips) + (1.75)(64.7 kips)
= 146.7 kips
Find the shear and moment capacity of the composite steel–concrete section.Find the plastic moment capacity, Mp, for interior girders. See Figure 2.32.
A App. D6.1; Tbl. D6.1-1
bs,ext effective flange width for exterior beam 87 in
bs,int effective flange width for interior beam 96 in
bc flange width, compression 9 in
D clear distance between flanges 22.54 in
Fyc specified minimum yield strength of the
compression flange 60 kips/in2
Fyt specified minimum yield strength of the
tension flange 60 kips/in2
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Fyw specified minimum yield strength of the web 60 kips/in2
tc = tf flange thickness, compression 0.68 ints thickness of the slab 8 intt = tf flange thickness, tension flange 0.68 intw web thickness 0.44 inCrt distance from top of concrete deck to top layer of longitudinal concrete deck reinforcement 3 inCrb distance from top of concrete deck to bottom layer of longitudinal concrete deck reinforcement 5 inPrb plastic force in the bottom layer of longitudinal deck reinforcementPrt plastic force in the top layer of longitudinal deck reinforcement
Locate the plastic neutral axis (PNA).
A App. D6-1; Tbl. D6.1-1; D6.1-2
2 in
Assume longitudinal reinforcement as shown.
Interior beam cross section
3 in3 in8 in 2 in
bf = 9 in
d = 23.9 in
2 in
D = 22.54 in
t?? = 8 in
tw = 0.44 in
tf and tc = 0.68 in
b? = 96 in
FIGURE 2.32Composite steel–concrete section for shear and moment capacity calculation.
89Design Examples
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The plastic compressive force in the slab is
Ps = 0.85 f′c bsts
= (0.85)(4.5 kips/in2)(96 in)(8 in)
= 2937.6 kips
Neglect strength of slab reinforcement.
Prb = Prt = 0 [conservative]
The plastic force in the compression flange is
Pc = Fycbctc = (60 ksi)(9 in)(0.68 in) = 367.2 kips
The plastic force in the web is
Pw = FywDtw = (60 ksi)(22.54 in)(0.44 in) = 595 kips
The plastic force in the tension flange is
Pt = Fytbttt = (60 ksi)(9 in)(0.68 in) = 367.2 kipsCase I (in web): Pt + Pw ≥ Pc + Ps
Pt + Pw = 367.2 kips + 595 kips = 962.2 kips
Pc + Ps = 367.2 kips + 2937.6 kips = 3305 kips
962.2 kips < 3305 kips [no good]
Case II (in top flange): Pt + Pw + Pc ≥ Ps
A App. Tbl. D6.1-1
Pt + Pw + Pc = 367.2 kips + 595 kips + 367.2 kips = 1329.4 kips
Ps = 2937.6 kips
1329.4 kips < 2937.6 kips [no good]
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© 2008 Taylor & Francis Group, LLC
Case III (in deck, below bottom reinforcement): Pt + Pw + Pc ≥ (Crb/ts)Ps
Pt + Pw + Pc = 367.2 kips + 595 kips + 367.2 kips = 1329.4 kips
(Crb/ts)Ps = (5 in/8 in)(2937.6) = 1836 kips
1329.4 kips < 1836 kips [no good]
Case IV (not considered because the effects of slab reinforcement are ignored: Prb = 0 and Prt = 0)
Case V (in deck, between reinforcement layers): Pt + Pw + Pc ≥ (Crb/ts)Ps
Pt + Pw + Pc = 367.2 kips + 595 kips + 367.2 kips = 1329.4 kips
(Crt/ts)Ps = (3 in/8 in)(2937.6) = 1102 kips
1329.4 kips > 1102 kips [OK]
The plastic neutral axis (PNA) is between the top and bottom layers of reinforcement, The PNA, Y , for Case V is,
A App. Tbl. D6.1-1
Y t P P P P PPs
rb c w t rt
s= + + + −
= ( ) + + + −8 0 367 2 595 367 2 0in kips kips kips kips ki. . ppskips2937 6.
= 3.62 in from top of deck
The distance from the mid-depth of steel compression flange to the PNA is
dc = 8 in + 2 in + (0.68 in)/2 – 3.62 in = 6.72 in
The distance from the mid-depth of steel web to the PNA is
dw = 8 in + 2 in + (23.9 in)/2 – 3.62 in = 18.33 in
The distance from the mid-depth of steel tension flange to the PNA is
dt = 8 in + 2 in + 23.9 in – (0.68 in)/2 – 3.62 in = 29.94 in
91Design Examples
© 2008 Taylor & Francis Group, LLC
See Figure 2.33.
The plastic moment capacity, Mp, for Case V in interior girders isA Tbl. D6.1-1
M Y Pt
P d P d P d P d Pps
srt rt rb rb c c w w t=
+ + + + +2
2ddt( )
= ( ) ( )
( )
+
+ +3 62 2937 6
2 8
0 0 3672. .
.in kips
in
22 6 72
631 18 33
367 2
kips in
kips in
ki
( )( )+( )( )+
.
.
. pps in( )( )
29 94.
= 2406 in-kips + 2467.6 in-kips + 11566.2 in-kips + 10994 in-kips
= 27,432 in-kips(1 ft/12 in)
= 2286 ft-kips
NOT E: Neglect strength of slab reinforcement (Prt = Prb = 0).
PNA
0.68 in
c.g.
Bottomflange
Section and cross section of interior girder
showing plastic neutral axis location
Web
0.68 in
0.68 in – 0.68 in = 11.27 in23.9 in
8 in
3.62 in = Y
PNA
–
22 in
SlabHaunch
Top flange
8 in
2 in
PNA
0.68 in
W24 × 78
9 in
Pt
Pw
Pc
dc
c.g.
8 in 2 in
dwdt
dt
dc
23.9 in
3.62 in = Y–
D
9 in
dtdtDt
dc
dw
FIGURE 2.33Section and cross section of interior girder for plastic moment capacity.
92 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Find the plastic moment capacity, Mp, for Case V in exterior girders. Please see Figure 2.34.
bs,ext effective flange width for exterior beam 87 in
bc flange width, compression 9 in
bt flange width, tension 9 in
D clear distance between flanges 22.54 in
Fyc specified minimum yield strength of the
compression flange 60 kips/in2
Fyt specified minimum yield strength of the
tension flange 60 kips/in2
Fyw specified minimum yield strength of the web 60 kips/in2
P total nominal compressive force in the concrete
deck for the design of the shear connectors at the
strength limit state kips
tc = tf flange thickness, compression 0.68 in
ts thickness of the slab 8 in
tt = tf flange thickness, tension flange 0.68 in
tw web thickness 0.44 in
Crt distance from top of concrete deck to top layer
of longitudinal concrete deck reinforcement 3 in
Crb distance from top of concrete deck to bottom layer
of longitudinal concrete deck reinforcement 5 in
Locate the plastic neutral axis (PNA).
A App. Tbl. D6.1-1
9 in
tw = 0.44 in
tf = 0.68 in
d = 23.9 in
2 in8 in
D = 22.54 in
bs = 87 in
FIGURE 2.34Composite cross section for exterior beam.
93Design Examples
© 2008 Taylor & Francis Group, LLC
The plastic compressive force in the slab is
Ps = 0.85 f′c bsts
= (0.85)(4.5 kips/in2)(87 in)(8 in)
= 2662.2 kips
Neglect strength of slab reinforcement.
Prb = Prt = 0 [conservative]
The plastic force in the compression flange is
Pc = Fycbctc = (60 ksi)(9 in)(0.68 in) = 367.2 kips
The plastic force in the web is
Pw = FywDtw = (60 ksi)(22.54 in)(0.44 in) = 595 kips
The plastic force in the tension flange is
Pt = Fytbttt = (60 ksi)(9 in)(0.68 in) = 367.2 kips
Case I (in web): Pt + Pw ≥ Pc + Ps
A App. Tbl. D6.1-1
Pt + Pw = 367.2 kips + 595 kips = 962.2 kips
Pc + Ps = 367.2 kips + 2662.2 kips = 3029.4 kips
962.2 kips < 3029.4 kips [no good]
Case II (in top flange): Pt + Pw + Pc ≥ Ps
Pt + Pw + Pc = 367.2 kips + 595 kips + 367.2 kips = 1329.4 kips
Ps = 2662.2 kips
1329.4 kips < 2662.2 kips [no good]
Case III (in deck, below bottom reinforcement): Pt + Pw + Pc ≥ (Crb/ts)Ps
Pt + Pw + Pc = 367.2 kips + 595 kips + 367.2 kips = 1329.4 kips
(Crb/ts)Ps = (5 in/8 in)(2662.2) = 1663.9 kips
1329.4 kips < 1663.9 kips [no good]
94 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Case IV (not considered because the effects of slab reinforcement are ignored: Prb = 0 and Prt = 0)
Case V (in deck, between reinforcement layers): Pt + Pw + Pc ≥ (Crt/ts)Ps
Pt + Pw + Pc = 367.2 kips + 595 kips + 367.2 kips = 1329.4 kips
(Crt/ts)Ps = (3 in/8 in)(2937.6) = 998.3 kips
1329.4 kips > 998.3 kips [OK]
The PNA is between the top and bottom layers of reinforcement. The PNA, Y , for Case V is
A App. Tbl. D6.1-1
Y t P P P P PPs
rb c w t rt
s= + + + −
= ( ) + + + −8 0 367 2 595 367 2 0in kips kips kips kips ki. . ppskips2662 2.
= 4.0 in from top of deck (was 3.62 in for interior girder)
The distance from the mid-depth of steel compression flange to the PNA is
dc = 8 in + 2 in + (0.68 in)/2 – 4.0 in = 6.34 in
The distance from the mid-depth of steel web to the PNA is
dw = 8 in + 2 in + (23.9 in)/2 – 4.0 in = 17.95 in
The distance from the mid-depth of steel tension flange to the PNA is
dt = 8 in + 2 in + 23.9 in – (0.68 in)/2 – 4.0 in = 29.65 in
See Figure 2.35.The plastic moment capacity, Mp, for Case V in exterior girders is
A Tbl. D6.1-1
95Design Examples
© 2008 Taylor & Francis Group, LLC
M Y Pt
P d P d P d P d Pps
srt rt rb rb c c w w t=
+ + + + +2
2ddt( )
= ( ) ( )
( )
+
+ +4 0 2662 2
2 8
0 0 367 22. .
.in kips
in
kkips in
kips in
kip
( )( )+( )( )+
6 34
595 17 95
367 2
.
.
. ss in( )( )
29 56.
= 2796.8 in-kips + 2328.0 in-kips + 10680.3 in-kips + 10854.4 in-kips
= 26,659.5 in-kips(1 ft/12 in)
= 2221.6 ft-kips
Assume an unstiffened web.
A Art. 6.10.9.2
NOT E: Neglect strength of slab reinforcement (Prt = Prb = 0).
The plastic shear resistance of the webs for both interior and exterior gird-ers is
A Eq. 6.10.9.2-2
Vp = (0.58)FywDtw
= (0.58)(60 ksi)(22.54 in)(0.44 in)
= 345.2 kips
Pt
Pw
Pc
dt
tc = 8 indc
23.9 in
Y = 4.0 in PNA–
2 in
Web
Bottom flange
Haunchtop flange
PNA
dw
Slab
2
FIGURE 2.35Section and cross section of exterior girder for plastic moment capacity.
96 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Please see Table 2.7.
Check the LRFD equation for the Strength I Limit State.
A Art. 6.10.6.2
Check the flexure for interior girders.
Confirm that the section is compact in positive flexure.A Art. 6.10.6.2.2
Fy = 60 ksi < 70 ksi [OK]
Dt
ininw
= = <22 540 44
51 2 150..
. [OK]
A Art. 6.10.2.1.1
The depth of the girder web in compression at the plastic moment, Dcp, is zero because the PNA is in the slab.
A App. D6.3.2The compact composite sections shall satisfy:
Art. 6.10.6.2.2; Eq. 6.10.6.2.2-1
2 2 0
0 440 3 76
Dt in
EF
cp
w yc= ( )( ) = ≤
.. [OK]
The section qualifies as a compact composite section.
The distance from the top of concrete deck to the neutral axis of composite section, Dp (=Y), is
Dp = 3.62 in for interior girders
Dp = 4.0 in for exterior girders
The total depth of composite section, Dt, is
TABLE 2.7
Summary of Plastic Moment Capacity and Shear Force
Girder Location Mp, ft-kips Vp, kips
Interior 2286 345.2Exterior 2221 345.2
97Design Examples
© 2008 Taylor & Francis Group, LLC
Dt = 23.9 in + 2 in + 8 in = 33.9 in
0.1 Dt = (0.1)(33.9 in) = 3.39 in
Determine the nominal flexural resistance for interior girders, Mn.
A Art. 6.10.7.1.2
If Dp ≤ 0.1 Dt, then Mn = Mp. Because Dp = 3.62 in ≥ 0.1 Dt = 3.39 in, the nomi-nal flexural resistance is
A Eq. 6.10.7.1.2-2
M MDDn p
p
t= −
1 07 0 7. .
= ( ) −
2286 1 07 0 7 3 62
33 9f s in
int-kip . . .
.
= 2275.1 ft-kips
The flange lateral bending stress, fl, is assumed to be negligible.
A Comm. 6.10.1
The resistance factor for flexure, Φf, is 1.0.A Art. 6.5.4.2
Check the following.
At the strength limit state, the section shall satisfy:
A Art. 6.10.7.1, A Eq. 6.10.7.1-1
M f S Mu l xt f n+ ≤13
ϕ = + ( )1167 13
0f ps Sxtt-ki
= 1167 ft-kips
ϕf nM f ps= ( )( )1 0 2275 1. . t-ki
= 2275.1 ft-kips ≥ 1167 ft-kips [OK]
Check the flexure for exterior girders.
Confirm that the section is a compact composite section.
98 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
It is the same as the interior girders; therefore, the section qualifies as a compact section.
The distance from the top of concrete deck to the neutral axis of composite section, Dp (= Y), is 4.0 in.
The total depth of composite section, Dt, is 33.9 in.
Determine the nominal flexural resistance for exterior girders, Mn
A Art. 6.10.7.1.2
If Dp ≤ Dt, then Mn = Mp. Because Dp = 4.0 in > 0.1 Dt = 3.39 in; therefore, the nominal flexural resistance is
A Eq. 6.10.7.1.2-2
M MDDn p
p
t= −
1 07 0 7. .
= ( ) −
2264 1 07 0 7 4 033 9
f ips inin
t-k . . ..
= 2235.5 ft-kips
The flange lateral bending stress, fl, is assumed to be negligible.A Comm. 6.10.1
The resistance factor for flexure, Φf, is 1.0.A Art. 6.5.4.2
Check the following.
At the strength limit state, the section shall satisfy:A Art. 6.10.7.1, A Eq. 6.10.7.1-1
M f S Mu l xt f n+ ≤13
ϕ = + ( )1288 0 13
0. ft-kips Sxt
= 1288.0 ft-kips
ϕf nM f ips= ( )( )1 0 2235 5. . t-k
= 2235.5 ft-kips ≥ 1288.0 ft-kips [OK]
99Design Examples
© 2008 Taylor & Francis Group, LLC
Find the shear resistance.
The resistance factor for shear, Φv, is 1.0.
A Art. 6.5.4.2
The nominal shear resistance of the unstiffened webs is
A Art. 6.10.9.2; A Eq. 6.10.9.2-1
Vn = Vcr = CVp
The ratio of shear-buckling resistance to shear yield strength, C, is 1.0 if the following is true.
A Art. 6.10.9.3.2; Eq. 6.10.9.3.2-4
Dt
EkFw yw
≤ 1 12.
Dt
ininw
= =22 540 44
51 2..
.
The shear-buckling coefficient, k, is
A Eq. 6.10.9.3.2-7
k SdD
o
= +
5 2
which is assumed to be equal to 5.0 inasmuch as do, the transverse spacing, is large, i.e., no stiffeners.
1 12 1 1229 000 5
6055 1. .
,.Ek
Fksi
ksiyw= ( )( ) =
Because (D/tw) = 51.2 < 55.1, C is 1.0.
Vn = CVp = (1.0)Vp = Vp = 345.2 kips
A Eq. 6.10.9.2-1
100 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Check that the requirement is satisfied for interior girders.A Eq. 6.10.9.1-1; Art. 6.10.9.1
Vu ≤ ΦvVn
156.3 kips < (1.0)(345.2 kips) = 345.2 kips [OK]
Check that the requirement is satisfied for exterior girders.
Vu ≤ ΦvVn
146.7 kips < (1.0)(345.2 kips) = 345.2 kips [OK]
Check the Service II Limit State LRFD equation for permanent deformations.A Tbls. 3.4.1-1, 3.4.1-2, 6.10.4.2
Q = 1.0(DC) + 1.0(DW) + 1.30(LL + IM)
Calculate the flange stresses, ff, for interior girders due to Service II loads.
The top steel flange for the composite section due to Service II loads must satisfy the following:
A Eq. 6.10.4.2.2-1
ff ≤ 0.95 RhFyf
The hybrid factor, Rh, is 1.0.A Art. 6.10.1.10.1
The yield strength of the flange, Fyf, is 60 ksi (given).Transform the concrete deck to an equivalent steel area with modular ratio
between steel and concrete, n, of 8. Please see Figures 2.36, 2.37, and 2.38.
9 in
d = 23.9 in
n = 82 in
8 in
96 in
FIGURE 2.36Interior girder section prior to transformed area.
101Design Examples
© 2008 Taylor & Francis Group, LLC
23.9 in – (2)(0.68 in) = 22.54 in23.9 in
2 in
33.9 in
0.44 in
0.68 in
0.68 in
96 in = 12 in8
9 in
9 in = 1.125 in8 8 in
FIGURE 2.37Interior girder section after transformed area.
7.4 in5
12 in
4
3
2
1
9 in
0.68 in
0.68 in2 in
23.9 in
23.9 in + 2 in + 8 in = 33.9 in
0.44 in 22.54 in–y = 26.5 in
C = 2.6 in
8 inCentroidal
Axis
FIGURE 2.38Dimensions for transformed interior beam section.
102 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Calculate the dimensions for the transformed interior girder section.
A1 = (0.68 in)(9.0 in) = 6.12 in2
A2 = (0.44 in)(22.45 in) = 9.92 in2
A3 = (0.68 in)(9.0 in) = 6.12 in2
A4 = (1.125 in)(2.0 in) = 2.25 in2
A5 = (12 in)(8.0 in) = 96.0 in2
ΣA = A1 + A2 + A3 + A4 + A5
= 6.12 in2 + 9.92 in2 + 6.12 in2 + 2.25 in2 + 96 in2
= 120.4 in2
Calculate the products of A and Y (from beam bottom).
A Y in in in1 12 36 12 0 68
22 08= ( )
=. . .
A Y in in in2 22 39 92 22 54
2118 54= ( )
=. . .
A Y in in in in3 32 36 12 23 9 0 68
2144 2= ( ) −
=. . . .
A Y in in in in4 42 32 25 23 9 2
256 03= ( ) −
=. . .
A Y in in in in in5 52 396 23 9 2 8
22870 4= ( ) + +
=. .
ΣAY A Y A Y A Y A Y A Y= + + + +1 1 2 2 3 3 4 4 5 5
= + + + +2 08 118 54 144 2 56 03 2870 43 3 3 3. . . . .in in in in iin3
= 3191.5 in3
The centroid of transformed section from the bottom of beam, 5, is
y AyA
inin
in= = =ΣΣ
3191 5120 4
26 53
2.
..
from bottom of beam to the centroidal axis.
103Design Examples
© 2008 Taylor & Francis Group, LLC
The moments of inertia are calculated as follows.
I bh Ad= +3
2
12
Iin in
in in in1
329 0 68
126 12 26 5 0 68
2= ( )( ) + ( ) −
.. . .
=2
44188 4. in
Iin in
in in2
320 44 22 54
129 92 26 5 0 6= ( )( ) + ( ) −
. .. . . 88 22 54
22520 0
24in in in−
=. .
Iin in
in in in3
329 0 68
126 12 26 5 0 68= ( )( ) + ( ) − −
.. . . 222 54 0 68
253 1
24. . .in in in−
=
Iin in
in in in4
321 125 2
122 25 26 5 23 9= ( )( ) + ( ) −
.. . . −−( ) =1 6 52 4in in.
Iin in
in in in5
32 212 8
1296 7 6 4 1756= ( )( ) + ( ) −( ) =. .22 4in
Itotal = I1 + I2 + I3 + I4 + I5
= 4188.4 in4 + 2520.0 in4 + 53.1 in4 + 6.5 in4 + 1756.2 in4
= 8524.2 in4
The distance between y and the top of the top flange of the steel section is
c = 26.5 in – 23.9 in = 2.6 in
The section modulus at the steel top flange is
S Ic
inin
intop = = =8524 22 6
3278 54
3..
.
The moment for composite dead load per interior or exterior girder, MDC, is 223.6 ft-kips; the moment for wearing surface per girder, MDW, is 36.6 ft-kips; and the total moment per interior or exterior girder, MLL+IM, is 475.9 ft-kips.
104 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
The steel top flange stresses, ff, due to Service II loads are
f MS
f ps inft
DCDC
top= =
( )
223 6 12
3278
.
.
t-ki
550 823 2in
kipsin
= .
f MS
f ps inft
DWDW
top= =
( )
36 6 12
3278 5
.
.
t-ki
iinkipsin3 20 13= .
f MS
f ps inft
LL IMLL IM
top+
+= =( )
475 9 12. t-ki =
3278 51 743 2.
.in
kipsin
ff = 1.0 fDC + 1.0 fDW + 1.3 fLL+IM
A Tbl. 3.4.1-1
= (1.0)(0.82 kips/in2) + (1.0)(0.13 kips/in2) + (1.3)(1.74 kips/in2)
= 3.21 kips/in2
Confirm that the steel top flange of the composite section due to Service II loads satisfies the requirement.
A Art. 6.10.4.2, A Eq. 6.10.4.2.2-1
ff ≤ 0.95 RhFyf
The hybrid factor, Rh, is 1.0.A Art. 6.10.1.10.1
0.95 RhFyf = (0.95)(1.0)(60 kips/in2)
= 57 kips/in2 > ff = 3.21 kips/in2 [OK]
Calculate the bottom steel flange stresses for composite sections, satisfying the following.
The bottom flange shall satisfyA Art. 6.10.4.2, A Eq. 6.10.4.2.2-2
f f R Ffl
h yf+ ≤2
0 95.
105Design Examples
© 2008 Taylor & Francis Group, LLC
The flange lateral bending stress, fl, is negligible (0.0).
The distance from the extreme tension fiber to the neutral axis is c = Y = 26.5 in.
The section modulus of the bottom flange is
S Ic
inin
inbottom = = =8524 226 5
321 74
3..
.
The bottom flange steel stresses, ff, due to Service II loads are
f MS
f ips inft
DCDC
bottom= =
( )
223 6 12
32
. t-k
11 78 343
2
.. /
inkips in=
f MS
f ps inft
DWDW
bottom= =
( )
36 6 12
321
. t-ki
... /
71 363
2
inkips in=
f MS
ips inft
LL IMLL IM
bottom+
+= =( )
475 9 12. ft-k
=
321 717 753
2
.. /
inkips in
ff = 1.0 fDC + 1.0 fDW + 1.3 fLL+IM
A Tbl. 3.4.1-1
= (1.0)(8.34 kips/in2) + (1.0)(1.36 kips/in2) + (1.3)(17.75 kips/in2)
= 32.78 kips/in2
Confirm that the steel bottom flange of composite section due to Service II loads satisfies the requirement.
A Art. 6.10.4.2, A Eq. 6.10.4.2.2-2
f f R Ffl
h yf+ ≤2
0 95.
32 780
20 95 1 0 602
2
2. . .kipsin
kipsin kips
in+ ≤ ( )
32.78 kips/in2 < 57 kips/in2 [OK]
106 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Calculate the flange stresses for exterior girders. Please see Figure 2.39.A Art. 6.10.4.2
Transform the concrete deck area to an equivalent area of steel using a modular ratio between steel and concrete, n, of 8. Please see Figures 2.40 and 2.41.
22.54 in23.9 in
2 in
33.9 in
0.44 in
0.68 in
–y
0.68 in
8 in
87 in = 10.875 in8
9 in
1.125 in
98 in =
FIGURE 2.40Exterior girder section after transformed area.
9 in
2 in
8 in
87 in
FIGURE 2.39Exterior girder section prior to transformed area.
107Design Examples
© 2008 Taylor & Francis Group, LLC
Calculate the dimensions for the transformed exterior girder section.
A1 = (0.68 in)(9.0 in) = 6.12 in2
A2 = (0.44 in)(22.45 in) = 9.92 in2
A3 = (0.68 in)(9.0 in) = 6.12 in2
A4 = (1.125 in)(2.0 in) = 2.25 in2
A5 = (10.875 in)(8.0 in) = 87.0 in2
ΣA = A1 + A2 + A3 + A4 + A5
= 6.12 in2 + 9.92 in2 + 6.12 in2 + 2.25 in2 + 87 in2
= 111.4 in2
7.67 in5
10.875 in
4
3
2
1
9 in
0.68 in
0.68 in2 in
8 in
23.9 in
33.9 in
0.44 in 22.54 in –y = 26.23 in
C = 2.33 in
Centroidalaxis
FIGURE 2.41Dimensions for transformed exterior beam section.
108 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Calculate the products of A and Y (from beam bottom).
A Y in in in1 12 36 12 0 68
22 08= ( )
=. . .
A Y in in in2 22 39 92 22 54
2118 54= ( )
=. . .
A Y in in in in3 32 36 12 23 9 0 68
2144 2= ( ) −
=. . . .
A Y in in in in4 42 32 25 23 9 2
256 03= ( ) −
=. . .
A Y in in in in in5 52 387 23 9 2 8
22601 3= ( ) + +
=. .
ΣAY A Y A Y A Y A Y A Y= + + + +1 1 2 2 3 3 4 4 5 5
= + + + +2 08 118 54 144 2 56 03 2601 33 3 3 3. . . . .in in in in iin3
= 2922.15 in3
The centroid of transformed section from the bottom of beam, y , is
y AyA
inin
in= = =ΣΣ
2922 15111 4
26 233
2..
. from bottom of beam
The moments of inertia are calculated as follows.
I bh Ad= +3
2
12
Iin in
in in in1
329 0 68
126 12 26 23 0 68= ( )( ) + ( ) −
.. . .
224102 4
24
= . in
Iin in
in in2
320 44 22 54
129 92 26 23 0= ( )( ) + ( ) −
. .. . .668 22 54
22442 8
24in in in−
=. .
109Design Examples
© 2008 Taylor & Francis Group, LLC
Iin in
in in in3
329 0 68
126 12 26 23 0 68= ( )( ) + ( ) −
.. . . −− −
=22 54 0 682
43 92
4. . .in in in
Iin in
in in i4
321 125 2
122 25 26 23 23 9= ( )( ) + ( ) −
.. . . nn in in−( ) =1 4 72 4.
Iin in
in in in5
32 210 875 8
1287 8 67 4= ( )( ) + ( ) −( ) =
.. 11635 8 4. in
Itotal = I1 + I2 + I3 + I4 + I5
= 4102.4 in4 + 2442.8 in4 + 43.9 in4 + 4.7 in4 + 1635.8 in4 = 8230.0 in4
Calculate the top steel flange stresses for composite sections, satisfying the following:
ff ≤ 0.95 RhFyf
A Eq. 6.10.4.2-1
The hybrid factor, Rh, is 1.0.
The distance between y and the top of the top flange of the steel sec-tion is
c = 26.23 in – 23.9 in = 2.33 in
The section modulus for the top flange of the steel section is
S Ic
inin
intop = = =8230 02 33
3532 24
3..
.
The steel top flange stresses, ff, due to Service II loads are
f MS
f ips inft
DCDC
top= =
( )
223 6 12
3532
.
.
t-k
220 763 2in
kipsin
= .
f MS
ips inft
DWDW
top= =
( )
36 6 12
3532 2
.
.
ft-k
iinkipsin3 20 12= .
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f MS
f ps inft
LL IMLL IM
top+
+= =( )
544 9 12. t-ki =
3532 21 853 2..
inkipsin
ff = 1.0 fDC + 1.0 fDW + 1.3 fLL+IM
A Tbl. 3.4.1-1
= (1.0)(0.76 kips/in2) + (1.0)(0.12 kips/in2) + (1.3)(1.85 kips/in2)
= 3.29 kips/in2
Confirm that the steel top flange of composite section due to Service II loads satisfies the requirement.
A Art. 6.10.4.2
ff ≤ 0.95 RhFyf
A Eq. 6.10.4.2.2-1
0.95 RhFyf = (0.95)(1.0)(60 kips/in2)
= 57 kips/in2 > ff = 3.29 kips/in2 [OK]
Calculate the bottom steel flange stresses for composite sections due to Service II loads satisfying the following.
Flange shall satisfy,A Art. 6.10.4.2. A Eq. 6.10.4.2.2-2
f f R Ffl
h yf+ ≤2
0 95.
The flange lateral bending stress, fl, is negligible (0.0).The hybrid factor, Rh, is 1.0.
A Art. 6.10.1.10.1
The distance from the extreme tension fiber to the neutral axis is c = y = 26.23 in.
The section modulus of the bottom flange is
S Ic
inin
inbottom = = =8230 026 23
313 764
3..
.
111Design Examples
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The bottom flange steel stresses, ff, due to Service II loads are
f MS
f s inft
DCDC
bottom= =
( )
223 6 12
31
. t-kip
33 768 553
2
.. /
inkips in=
f MS
ips inft
DWDW
bottom= =
( )
36 6 12
313
. ft-k
... /
761 403
2
inkips in=
f MS
f ps inft
LL IMLL IM
bottom+
+= =( )
544 9 12. t-ki
=
313 7620 843
2
.. /
inkips in
ff = 1.0 fDC + 1.0 fDW + 1.3 fLL+IM
A Tbl. 3.4.1-1
= (1.0)(8.55 kips/in2) + (1.0)(1.40 kips/in2) + (1.3)(20.84 kips/in2)
= 37.04 kips/in2
Confirm that the steel bottom flange of composite section due to Service II loads satisfies the requirement.
A Art. 6.10.4.2, A Eq. 6.10.4.2.2-2
f f R Ffl
f h yf+ ≤2
0 95.
37 040
20 95 1 0 602
2
2. . .kipsin
kipsin kips
in+ ≤ ( )
37.04 kips/in2 < 57 kips/in2 [OK]
Check LRFD Fatigue Limit State II.
Details shall be investigated for the fatigue as specified in AASHTO Art. 6.6.1. The fatigue load combination in AASHTO Tbl. 3.4.1-1 and the fatigue load specified in AASHTO Art. 3.6.1.4 shall apply.
A Art. 6.6.1; 3.6.1.4
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Q = 0.75(LL + IM)A Tbl. 3.4.1-1
For load-induced fatigue, each detail shall satisfy,A Eq. 6.6.1.2.2-1; Art. 6.6.1.2
γ(Δf) ≤ (ΔF)n
The force effect, live load stress range due to fatigue load is (Δf). The load factor is γ. The nominal fatigue resistance is (ΔF)n.
A 6.6.1.2.2
The fatigue load is one design truck with a constant spacing of 30 ft between 32 kip axles.
A Art. 3.6.1.4.1
Find the moment due to fatigue load.
Please see Figure 2.42.Determine the moment due to fatigue load.
ΣM@B = 0
RA(40 ft) = (32 kips)(20 ft) + (8 kips)(6 ft)
RA = 17.2 kips
Mc = (17.2 kips)(20 ft) = 344 ft-kips
The dynamic load allowance, IM, is 15%.A Tbl. 3.6.2.1-1
C
20 ft20 ft
30 ft32 kips 32 kips 8 kips 6 ft14 ft
R
A B
FIGURE 2.42Single lane fatigue load placement with one design truck load for maximum moment at midspan.
113Design Examples
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The approximate load distribution factors for one traffic lane shall be used. Divide out the multiple presence factor, m, which was already included in the approximate equations for distribution factors, so it is 1.2.
Comm. 3.6.1.1.2; A Art. 3.6.1.4.3b
The load distribution factor for fatigue moment in interior girders with one design lane loaded, is
DFM DFMfat
si,int .
..
.1 2
0 4981 2
0 415= =
The load distribution factor for fatigue moment in exterior girders with one design lane loaded is
DFM DFMfat ext
se, .
..
.1 2
0 751 2
0 625= =
The unfactored, distributed fatigue moment is
Mfat = Mfat,LL = (DFMfat)(1 + IM)
The unfactored, distributed fatigue moment for interior girders is
Mfat,int = (344 ft-kips)(0.415)(1 + 0.15) = 164.2 ft-kips
The unfactored, distributed fatigue moment for exterior girders is
Mfat,ext = (344 ft-kips)(0.625)(1 + 0.15) = 247.3 ft-kips
Use Sbottom because this causes the maximum stress in either the top steel or the bottom steel flange.
The maximum stress due to fatigue load for interior girders is
∆f MS
f ips inftfat
bottomint
,int.
= =
164 2 12t-k =
321 76 123 2..
inkipsin
The maximum stress due to fatigue load for exterior girders is
∆f MS
f ips inft
extfat ext
bottom= =
,
.247 3 12t-k =
313 769 463 2.
.in
kipsin
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The girder is in Detail Category A, because it is a plain rolled member.
A Tbl. 6.6.1.2.3-1
A = 250 x 108 kips/in2 (constant value taken for Detail Category A)
A Tbl. 6.6.1.2.5-1
The number of cycles per truck passage, n, is 2.0 for simple span girders with span equal to and less than 40 ft.
A Tbl. 6.6.1.2.5-2
The fraction of truck traffic in a single lane, p, is 0.80 because the num-ber of lanes available to trucks, NL, is 3 lanes.
A Tbl. 3.6.1.4.2-1
The single lane daily truck traffic is averaged over the design life,
ADTTSL = (p)(ADTT) = (0.80)(2500) = 2000 trucks per day
A Eq. 3.6.1.4.2-1
The number of cycles of stress range for 75 years N is.
A Art. 6.6.1.2.5; Eq. 6.6.1.2.5-3
N = (365 days)(75 years)n(ADTT)SL = (47,625)(2.0)(2000) = 109,500,000 cycles
The constant amplitude fatigue threshold, (ΔF)TH, is 24.0 kips/in2 for Detail Category A.
A Tbl. 6.6.1.2.5-3
The nominal fatigue resistance, (ΔF)n, is
A Art. 6.6.1.2.2; Eq. 6.6.1.2.5-2
∆F ANn( ) =
1 3/
=
250 10
109 500 000
82
1 3
x kipsincycles, ,
/
= 6.11 kips/in2
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Confirm that the following formula is satisfied.A Art. 6.6.1.2.2
γ(Δf) ≤ (ΔF)n
where:γ = load factor in Table 3.4.1-1 for Fatigue II load combination(Δf) = force effect, live load stress range due to the passage of the fatigue
load (Art. 3.6.1.4)
The factored live load stress due to fatigue load for interior girders isA Tbl. 3.4.1-1
γ(Δfint) = (0.75)(LL + IM)(Δfint) = (0.75)(1.0)(6.12 kips/in2)
= 4.59 kips/in2 < (ΔF)n = 6.11 kips/in2 [OK]
The factored live load stress due to fatigue load for exterior girders is
γ(Δfext) = (0.75)(LL + IM)(Δfext) = (0.75)(1.0)(9.46 kips/in2)
= 7.09 kips/in2 > (ΔF)n = 6.11 kips/in2 [NG]
To control web-buckling and elastic flexing of the web, provisions of AASHTO Art. 6.10.5.3 shall be satisfied for distortion-induced fatigue.
A Art. 6.10.5.3
The ratio of the shear buckling resistance to the shear specified mini-mum yield, C, is 1.0 (see previous calculations).
A Eq. 6.10.9.3.2-4
The plastic shear force, Vp, was 345.2 kips for both interior and exterior girders.
The shear buckling resistance, Vcr, isA Eq. 6.10.9.3.3-1
Vn = Vcr = CVp = (1.0)(345.2 kips) = 345.2 kips
Find the shear due to factored fatigue load. Please see Figure 2.43.
ΣM@B = 0
R kips ftft
kips kipsA = +
( ) =32 1040
32 40
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The shear due to fatigue load, V, is 40 kips.
The distribution factor for shear due to fatigue load for interior girders with one design lane loaded is
DFV DFVfat,int
int
...
.= = =1 2
0 681 2
0 57
The distribution factor for shear due to fatigue load for exterior girders with one design lane loaded is
DFV DFVfat ext
ext, .
..
.= = =1 2
0 751 2
0 625
The shear for the total component dead load for interior girders, VDC, is 22.4 kips; and the shear for the wearing surface dead load for interior girders, VDW, is 3.66 kips.
Special Fatigue Requirements for WebsA Art. 6.10.5.3
The factored fatigue load shall be taken as twice that calculated using the fatigue load combination specified in AASHTO Tbl. 3.4.1-1, with the fatigue live load taken as in AASHTO 3.6.1.4.
Interior panels of web shall satisfy
Vu ≤ Vcr
A Eq. 6.10.5.3-1
where:
40 ft
32 kips 32 kips30 ft 10 ft
R
A B
FIGURE 2.43Single lane fatigue load placement with one design truck load for maximum shear at support.
117Design Examples
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Vu = shear in the web due to the unfactored permanent load plus the factored fatigue load
Vcr = shear-buckling resistance determined from AASHTO Eq. 6.10.9.3.3-1
= 345.2 kips (previously calculated)Vpermanent = VDC + VDW
= 22.44 kips + 3.66 kips (previously calculated) = 26.1 kips
The factored fatigue load is
γ = load factor = 0.75 for Fatigue II loadA Tbl. 3.4.1-1
IM = dynamic allowance for fatigue = 15%A Tbl. 3.6.2.1-1
The factored fatigue for girders is given as
Vf = 2(DFV)(γ)(V)(LL + IM)
For interior beams the factored fatigue load is
Vf,int = (2)(DFVint)(γ)(V)(LL + IM)
= (2)(0.57)(0.75)(40.0 kips)(1 + 0.15)
= 39.33 kips for interior beams
For exterior beams the factored fatigue load is
Vf,ext = (2)(DFVext)(γ)(V)(LL + IM)
= (2)(0.625)(0.75)(40.0 kips)(1 + 0.15)
= 43.13 kips for exterior beams [controls]
Vu = Vpermanent + Vf
= 26.1 kips + 43.13 kips = 69.23 kips
Interior panel webs shall satisfy:
Vu ≤ Vcr
A Art. 6.10.5.3; Eq. 6.10.5.3-1
Vu = 69.23 kips ≤ Vcr = 345.2 kips [OK]
It is noted that the application of Art. 6.10.5.3 for Fatigue I load is considered conservative for Fatigue II load.
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Design Example 4: Longitudinal Steel Girder
Situation
A simple span noncomposite steel girder rural highway bridge with a span of 40 ft is shown in Figure 2.44.
The overall width of the bridge is 33 ft 4 in.The clear (roadway) width is 28 ft 0 in.The roadway is a concrete slab 7 in thick (dead load of 145 lbf/ft3, f′c = 4
kips/in2) supported by four W33 x 130, A992 Grade 50 steel girders that are spaced at 8 ft 4 in apart.
The compression flange is continuously supported by the concrete slab, and additional bracing is provided at the ends and at midspan.
Noncomposite construction is assumed.There is a wearing surface of 3 in thick bituminous pavement (dead
load of 0.140 kips/ft3).The barrier, sidewalk, and railings combine for a dead load of 1 kip/ft.
Elevation
33 ft 4 in
28 ft
4 ft 2 in 4 ft 2 in3 at 8 ft 4 in = 25 ft 0 in
4 girders: W33 × 130
2 ft 8 in2 ft 8 in
L = 40 ftA
A
7 in slab
FIGURE 2.44Steel girder bridge, 40 ft span.
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Requirements
Review the longitudinal girders for adequacy against maximum live load, dead load, and shear load for load combination limit states Strength I, Service I, and Fatigue II. Use the AASHTO HL-93 loading, and assume an urban highway with 8,000 vehicles per lane per day.
Solution
For each limit state, the following limit state equation must be satisfied.A Art. 1.3.2
ΣηDηRηIγiQi ≤ ΦRn = Rr
The following terms are defined for limit states.A Eq. 1.3.2.1-1
Q = U = factored load effect (U = Σ ηiγiQi ≤ ΦRn = Rr)Qi force effect from various loadsRn nominal resistanceRr factored resistance (ΦRn)γi load factor; a statistically based multiplier applied to force effects
including distribution factors and load combination factorsηi load modifier relating to ductility, redundancy, and operational importance 1.0 (for conventional designs)
A Art. 1.3.3
ηD ductility factor (strength only) 1.0 (for conventional designs)A Art. 1.3.4
ηR redundancy factor (strength only) 1.0 (for conventional designs)A Art. 1.3.5
ηI operational importance factor (strength and extreme only) 1.0 (for conventional designs)Φ resistance factor, a statistically based multiplier applied to nominal resistance 1.0 (for Service Art. 1.3.2.1 and Fatigue Limit Art. C6.6.1.2.2 States)Φ resistance factor For concrete structures see A Art. 5.5.4.2.1 For steel structures see A Art. 6.5.4.2
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Step 1: Select Load Modifiers, Combinations, and Factors
See Table 2.8.A Arts. 1.3.3, 1.3.4, 1.3.5, 1.3.2.1; Eq. 1.3.2.1-2
Load Combinations and FactorsA Art. 3.4.1
Q = Ü is the total factored force effect, Σ ηiγiQi.Qi represents the force effects from loads specified.η represents specified load modifier.γi represents specified load factors.
The following load combinations are used in this example.A Tbls. 3.4.1-1, 3.4.1-2
Strength I Limit State: Q = (1.25 DC + 1.50 DW + 1.75(LL + IM))Service I Limit State: Q = (1.0(DC + DW) + 1.0(LL + IM))Fatigue II Limit State: Q = (0.75)(LL + IM)
Step 2: Determine Maximum Live Load Moments at Midspan
Please see Figures 2.45a–e.A Art. 3.6.1.2
The HL-93 live load consists of the design lane load and either the design tandem or the design truck load (whichever is greater). The design truck is HS-20. The design tandem consists of a pair of 25.0 kip axles, spaced at 4.0 ft apart, and a transverse spacing of wheels is 6.0 ft. The design lane load consists of 0.64 kips/ft uniformly distributed in the longitudinal direc-tion. Transversely, the design lane load is distributed uniformly over a 10.0 ft width, within a 12.0 ft design lane.
The force effects from the design lane load are not subject to a dynamic load allowance.
A Art. 3.6.2.1
The fatigue truck (HS-20) is placed in a single lane with a constant spacing of 30 ft between rear axles. See Figure 2.45e.
A Art. 3.6.1.4.1
TABLE 2.8
Load Modifier Factors
Load Modifier Strength Service Fatigue
Ductility, ηD 1.0 1.0 1.0
Redundancy, ηr 1.0 1.0 1.0
Operational importance, ηI 1.0 N/A N/A
ηi = ηDηRηI ≥ 0.95 1.0 1.0 1.0
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20 ft20 ft
6 ft
3 3
810
6 ft4 ft
FIGURE 2.45aInfluence line for maximum moment at midspan.
40 ft
CL
6 ft
Truck
32 kips 32 kips 8 kips
6 ft14 ft 14 ft
A B
FIGURE 2.45bControlling load position for moment at midspan for design truck load (HS-20).
CLTandem
25 kips 25 kips
20 ft 16 ft4 ft
A B
FIGURE 2.45cControlling load position for moment at midspan for design tandem load.
Lane
w = 0.64 kips/ft
A B
FIGURE 2.45dControlling load position for moment at midspan for design lane load.
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Using the influence line diagrams (IL), Figure 2.45a
Mtr = (32 kips)(10 ft) + (32 kips + 8 kips)(3 ft) = 440 ft-kips per lane
Mtandem = (25 kips)(10 ft + 8 ft) = 450 ft-kips per lane [controls]
M wLkips
ftft
f ipln
.= =
( )
=2
2
8
0 64 40
8128 t-k ss
per lane
Mfatigue = (32 kips)(10 ft) + (8 kips)(3 ft) = 344 ft-kips per lane
Step 3: Calculate Live Load Force Effects for Moment, Qi
Where w is the clear roadway width between curbs and/or barriers, the number of design lanes is,
A Art. 3.6.1.1.1
N w ftft
lane
lanesL = = =12
28
122 33. (2 lanes)
The modulus of elasticity of concrete where wc = 0.145 kips/in3 is
A Eq. 5.4.2.4-1
E w fc c c= ( )( ) ′33 000 1 5, .
= ( )
33 000 0 145 43
1 5
2, ..kips
inkipsin
= 3644 kips/in2
40 ft
CLFatigueloading
32 kips 32 kips 8 kips
6 ft30 ft 14 ft
A B
FIGURE 2.45eSingle lane fatigue load placement with one design truck load for maximum moment at mid-span. (A Art. 3.6.1.4.1)
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The modular ratio between steel and concrete is
n EE
kipsin
kipsin
s
c= =
29 000
3644
2
2
,
= 7.96 (we will use 8)
n modular ratio between steel and concrete 8AISC Tbl. 1-1 for W 33 × 130
d depth of beam 33.10 inA area of beam 38.3 in2
I moment of inertia of beam 6710 in4
Ebeam modulus of elasticity of beam 29,000 kips/in2
Edeck modulus of elasticity of concrete deck 3644 kips/in2
L span length 40 ftts slab thickness 7 inbf flange width 11.5 intf flange thickness 0.855 inSx section modulus 406 in3
tw web thickness 0.58 in
See Figure 2.46.
11.5 in
Sx for steel beam = 406 in3
W33 × 130
eg = 20.0 in tw = 0.58 in
D = 31.39 in d = 33.1 in
0.855 in
7 in slab
FIGURE 2.46Noncomposite steel section at midspan.
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The distance between the centers of the slab and steel beam is
e d t in in ings= + = + =
2 233 1
27
220 0. .
The longitudinal stiffness parameter, Kg, is
A Art. 4.6.2.2.1, A Eq. 4.6.2.2.1-1
Kg = n(I + Aeg2) = (8)(6710 in4 + (38.3 in2)(20 in)2) = 176,240 in4
KLt
inft in
g
s12176 240
12 40 71 03
4
3=( )( )( )
≈, .
Type of deck is (a).
A Tbl. 4.6.2.2.1-1
For interior beams with concrete decks, the live load moment may be determined by applying the lane fraction specified in AASHTO Table 4.6.2.2.2b-1, or Appendix A.
A Art. 4.6.2.2.2b
The multiple presence factor, m, applies to the lever rule case for live load distribution factors only.
A Art. 3.6.1.1.2
In the approximate equations for distribution factors, multiple factors are already included.
The distribution of live load moment per lane for interior beams of typ-ical deck cross section (a) with one design lane loaded is
A Tbl. 4.6.2.2b-1 or Appendix A
DFM S SL
KLtsig
s= +
0 06
14 12
0 4 0 3
3.. .
0 1.
= +
0 06 8 3314
8 3340
10 4 0 3
. . .. .ft ft
ft.. .0 0 1( )
= 0.567 lanes
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The distribution of live load moment per lane for interior beams with two or more design lanes loaded is
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
= +
0 075 8 339 5
8 3340
0 6 0
. ..
.. .ft ft
ft
220 11 0. .( )
= 0.750 lanes [controls]
For exterior beams with type (a) concrete decks, the live load moment may be determined by applying the lane fraction specified in Appendix B.
A Art. 4.6.2.2.2d; Tbl. 4.6.2.2.2d-1
Use the lever rule to find the distribution factor for moments for exte-rior beams with one design lane loaded. See Figure 2.47.
ΣM@b = 0
Ra(100 in) = (0.5 P)(22 in) + (0.5 P)(94 in)
Ra = 0.58 P = 0.58
100 in
P/2 P/2
72 in24 in
18 in 22 in
+ba
32 in
Assume hinge
FIGURE 2.47Lever rule for determination of distribution factor for moment in exterior beam, one lane loaded.
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The distribution of the live load moment per lane for exterior beams of deck cross section type (a) with one design lane loaded is
A Tbl. 4.6.2.2.2d-1 or Appendix B
DFMse = (multiple presence factor for one loaded lane) (Ra) = (1.2)(0.58) = 0.696
A Tbl. 3.6.1.1.2-1
Find the distribution factor for moments for exterior girders with two of more design lanes loaded, DFMme, using g, an AASHTO distribu-tion factor Table 4.6.2.2.2d-1 or Appendix B.
g = (e)(ginterior)
ginterior = DFMmi = 0.750
The distance from the exterior web of the exterior beam to the interior edge of the curb or traffic barrier, de, is 18 in (1.5 ft).
The correction factor for distribution is
e d fte= + = + =0 779 1
0 77 1 59 1
0 935..
. ..
.
The approximate load distribution factor for moments for exterior gird-ers with two or more design lanes loaded has the multiple presence factors included.
A Art. 3.6.1.1.2
DFMme = (e)(DFMmi) = (0.935)(0.750) = 0.701
Step 4: Determine Maximum Live Load Shears
Please see Figures 2.48a–e.
Using the influence line diagram, Figure 2.48a
Vtr = (32 kips)(1 + 0.65) + (8 kips)(0.3) = 55.2 kips [controls]
Vtandem = (25 kips)(1 + 0.9) = 47.5 kips
V wLkips
ftft
kipsln
..= =
( )
=2
0 64 40
212 8 per lane
Vfatigue = (32 kips)(1 + 0.25) = 40.0 kips
127Design Examples
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40 ft
14 ft 12 ft
10 ft
0.250.3
0.65
0.91.0
A
14 ft
4 ft
FIGURE 2.48aMaximum live load shears; influence line for maximum shear at support.
40 ft
32 kips 32 kips 8 kips
12 ft14 ft14 ft
A B
FIGURE 2.48bControlling load position for shear at support for design truck load (HS-20).
4 ft36 ft
25 kips25 kips
A B
FIGURE 2.48cControlling load position for shear at support for design tandem load.
w = 0.64 kips/ft
A B
FIGURE 2.48dControlling load position for shear at support for design lane load.
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Step 5: Calculate the Live Load Force Effects for Shear, Qi
A Arts. 4.6.2.2.3a and 3b; Tbls. 4.6.2.2.3a-1 and 3b-1
For interior beams with concrete decks, the live load shears may be deter-mined by applying the lane fraction specified in Table 4.6.2.2.3a-1 or Appendix C.
The distribution for the live load shears for interior beams with one design lane loaded is
DFV S ftsi = +
= +
=0 3625
0 36 8 3325
0. . . .6693 lanes
The distribution for the live load shears for interior beams with two or more design lanes loaded is
DFV S S ftmi = +
−
= +0 212 35
0 2 8 3312
2
. . .
−
=8 3335
0 8382. .ft lanes [controls]
A Art. 4.6.2.2.3a; Tbl. 4.6.2.2.3a-1 or Appendix C
For exterior beams with type (a) concrete decks, the live load shears may be determined by applying the lane fraction in Table 4.6.2.2.3b-1 or Appendix D.
Using the lever rule, the distribution factor for live load shear for exte-rior beams with one design lane loaded is the same as the corre-sponding live load moment.
Therefore, DFVse = DFMse = 0.696 lanes.
Find the distribution factor for live load shear for exterior beams with two or more design lanes loaded. The correction factor for distribu-tion is
e d fte= +
= +
=0 610
0 6 1 510
0 75. . . .
DFVmi = g = (e)(ginterior) = (e)(DFVmi) = (0.75)(0.838) = 0.629 lanes
Please see Table 2.9.
8 kips 32 kips 32 kips
14 ftA B
10 ft30 ft
FIGURE 2.48eSingle lane fatigue load placement with one design truck load for maximum shear at support.
129Design Examples
© 2008 Taylor & Francis Group, LLC
When a bridge is analyzed for fatigue load using the approximate load distribution factors, one traffic lane shall be used. Therefore, the force effect shall be divided by 1.20.
Please see Table 2.10.A Art. 3.6.1.1.2, 3.6.1.4.3b
Step 6: Calculate the Live Load Moments and Shears
The distributed live load moment per interior girder with multiple (two or more) lanes loaded governs is
A Art. 3.6.2.1
MLL+IM = DFMint((Mtr or Mtandem)(1 + IM) + Mln)
= (0.75)((450 ft-kips)(1 + 0.33) + 128 ft-kips)
= 545 ft-kips
The distributed live load moment per exterior girder with multiple lanes loaded is
MLL+IM = DFMext((Mtr or Mtandem)(1 + IM) + Mln)
= (0.701)((450 ft-kips)(1 + 0.33) + 128 ft-kips)
= 509.3 ft-kips
TABLE 2.9
Summary of Distribution Factors:
Load Case DFMint DFMext DFVint DFVext
Distribution Factors from A Art. 4.6.2.2.2
Multiple Lanes Loaded 0.750 0.701 0.838 0.629Single Lane Loaded 0.567 0.696 0.693 0.696
Design Value 0.750 0.701 0.838 0.696
TABLE 2.10
Summary of Fatigue Limit State Distribution Factorsa
Load Case DFMint DFMext DFVint DFVext
Distribution Factors from A Art. 4.6.2.2.2
Multiple Lanes Loaded N/A N/A N/A N/ASingle Lane Loaded 0.567/1.2
= 0.4730.696/1.2 = 0.580
0.693/1.2 = 0.578
0.696/1.2 = 0.580
Design Value 0.473 0.580 0.578 0.580
a Divide values above by multiple presence factor of 1.20.
130 Simplified LRFD Bridge Design
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A reduced dynamic load allowance of 15% is applied to the fatigue load.
A Tbl. 3.6.2.1-1
For a single lane loaded, the fatigue moment for interior beams is
M DFMm
M LL IMfatigue IMsi
fatigue+ = +( )
=
( )( )0 567
1 2344 1 15.
..f pst-ki
= 187.0 ft-kips
For a single lane loaded, the fatigue moment for exterior beams is
M DFMm
M LL IMfatigue IMse
fatigue+ = +( )
=
( )( )0 696
1 2344 1 15.
..f ipst-k
= 229.0 ft-kips per beam
The distributed live load shear for interior beams is,
VLL+IM = DFV((Vtr or Vtandem)(1 + IM) + Vln)
The distributed live load shear for interior beams (DFVmi = 0.838) is
VLL+IM = (0.838)((55.2 kips)(1 + 0.33) + 12.8 kips) = 72.2 kips
The distributed live load shear for exterior beams (DFVse = 0.696) is
VLL+IM = (0.696)((55.2 kips)(1 + 0.33) + 12.8 kips) = 60.0 kips
The fatigue shear for interior beams is
V DFVm
V LL IMfatigue IMsi
fatigue+ = +( )
131Design Examples
© 2008 Taylor & Francis Group, LLC
=
( )( )0 693
1 240 1 15.
..kips
= 26.6 kips
The fatigue shear for exterior beams is,
V DFVm
V LL IMfatigue IMse
fatigue+ = +( )
=
( )( )0 696
1 240 1 15.
..kips
= 26.7 kips
Step 7: Calculate Force Effects from the Dead Load and Wearing Surface
Find the force effects from the dead load and wearing surface for interior beams.The nominal weight of the deck slab is
w kipsft
in ftinslab, int .=
( )
0 145 7 1
123
( ) =8 33 0 705. . /ft kips ft
The nominal weight of a W33 x 130 beam, w33x130, is 0.130 kips/ft.
wDC = 0.705 kips/ft + 0.130 kip/ft = 0.835 kips/ft
The nominal weight of the wearing surface is,
w kipsft
in ftinDW =
( )
0 140 3 112
8 33. . 33 0 292ft kips ft( ) = . /
The interior beam moments and shears are,
V w Lkips
ftft
kipsDCDC= =
( )
=2
0 835 40
216 7
..
M w Lkips
ftft
fDCDC= =
( )
=2
2
8
0 835 40
8167
.t-kkips
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V w Lkips
ftft
kipsDWDW= =
( )
=2
0 292 40
25 84
..
M w Lkips
ftft
fDWDW= =
( )
=2
2
8
0 292 40
858 4
.. t--kips
Table 2.11 summarizes the loads, shears, and moments for interior beams.
Find the force effects from the dead load and wearing surface for exterior beams.The nominal weight of the deck slab is
w kipsft
in ftinslab ext, .=
( )
0 145 7 1
123
+
=8 332
1 5 0 479. . . /ft ft kips ft
The nominal weight of a W33 x 130 beam, wW33x130, is 0.130 kips/ft. Assume that the nominal weight of the barrier, sidewalk, and railings, wbarrier+sidewalk+rail, is 1.0 kips/ft. For exterior girders, the distributed load for the slab, beam, bar-rier, sidewalk, and girders is
wDC = wslab,ext + wW33x130 + wbarrier+sidewalk+rail
= 0.479 kips/ft + 0.130 kips/ft + 1.0 kips/ft = 1.61 kips/ft
The nominal weight of the wearing surface is
w kipsft
in ftinDW =
( )
0 140 3 112
8 33. . 33
21 5 0 198ft ft kips ft+
=. . /
The exterior beam moments and shears are
V w Lkips
ftft
kipsDCDC= =
( )
=2
1 61 40
232 2
..
TABLE 2.11
Summary of Loads, Shears, and Moments in Interior Beams
Load Typew
(kip/ft)Moment(ft-kips)
Shear(kips)
DC 0.835 167 16.7DW 0.292 58.4 5.84LL + IM N/A 545 72.2Fatigue + IM N/A 187 26.6
133Design Examples
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M w Lkips
ftft
DCDC= =
( )
=2
2
8
1 61 40
8322
.ft-kiips
V w Lkips
ftft
kipsDWDW= =
( )
=2
0 198 40
23 96
..
M w Lkips
ftft
DWDW= =
( )
=2
2
8
0 198 40
839 6
.. ft--kips
Table 2.12 summarizes the loads, shears, and moments for exterior beams.
Step 8: Find the Factored Moments and Shears with Applicable Limit States
A Tbls. 3.4.1-1, 3.4.1-2
Strength I Limit State for Interior Beam
U = (1.25 DC + 1.50 DW + 1.75(LL + IM))
Vu = ((1.25)(16.7 kips) + (1.50)(5.84 kips) + (1.75)(72.2 kips))
= 156 kips
Mu = ((1.25)(167 ft-kips) + (1.50)(58.4 ft-kips) + (1.75)(545 kips))
= 1250 ft-kips
Service I Limit State
U = (1.0(DC + DW) + 1.0(LL + IM))
TABLE 2.12
Summary of Loads, Shears, and Moments in Exterior Beams
Load Typew
(kip/ft)Moment(ft-kips)
Shear(kips)
DC 1.61 322 32.2DW 0.198 39.6 3.96LL + IM N/A 509.3 60.0Fatigue + IM N/A 229 26.7
134 Simplified LRFD Bridge Design
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Vu = ((1.0)(16.7 kips + 5.84 kips) + (1.00)(72.2 kips))
= 94.74 kips
Mu = ((1.0)(167 ft-kips + 58.4 ft-kips) + (1.00)(545 kips))
= 770.4 ft-kips
Fatigue II Limit State for Interior Beams (with dead loads considered to be conservative)
U = η(1.0 DC + 1.0 DW + 0.75(LL + IM))
Vu = (1.0)((1.0)(16.7 kips) + (1.0)(5.84 kips) + (0.75)(26.6 kips))
= 42.5 kips
Mu = (1.0)((1.0)(167 ft-kips) + (1.0)(58.4 ft-kips) + (0.75)(187 kips))
= 366 ft-kips
Strength I Limit State for Exterior BeamA Tbls. 3.4.1-1, 3.4.1-2
U = (1.25 DC + 1.50 DW + 1.75(LL + IM))
Vu = ((1.25)(32.2 kips) + (1.50)(3.96 kips) + (1.75)(60.0 kips))
= 151.0 kips
Mu = ((1.25)(322 ft-kips) + (1.50)(39.6 ft-kips) + (1.75)(509.3 kips))
= 1353.2 ft-kips [controls]
Service I Limit State
U = (1.0(DC + DW) + 1.0(LL + IM))
Vu = ((1.0)(32.2 kips + 3.96 kips) + (1.00)(60.0 kips))
= 96.2 kips
Mu = ((1.0)(322 ft-kips + 39.6 ft-kips) + (1.00)(509.3 kips))
= 870.9 ft-kips
135Design Examples
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Fatigue II Limit State for Exterior Beams (with dead load considered to be conservative)
U = (1.0 DC + 1.0 DW + 0.75(LL + IM))
Vu = ((1.0)(32.2 kips) + (1.0)(3.96 kips) + (0.75)(26.7 kips))
= 56.2 kips
Mu = ((1.0)(322 ft-kips) + (1.0)(39.6 ft-kips) + (0.75)(229 kips))
= 533.4 ft-kips
Step 9: Check Fundamental Section Properties for Strength Limit I
For the Strength I Limit State, the exterior beam moment controls (please refer to Step 7):
The maximum factored moment, Mu, is 1353.2 ft-kips.
For noncomposite sections,
A Art. C6.10.1.2; A App. A6.1.1, A6.1.2; Art. 6.5.4.2
Mu ≤ ΦfMn
Because Mn is the required nominal moment resistance and Φf is 1.0 for steel structures, Mn must be at least Mu = 1353.2 ft-kips.
Check the plastic section modulus required.
Z MF
f ips inft
kireq du
y’
t-k≥ =
( )
1353 2 12
50
.
ppsin
in2
3324 7= .
ZW33x130 = 467 in3 ≥ 324.7 in3 [OK]
AISC Tbl.1-1
Check the web proportions.
A Art. 6.10.2.1.1, A Eq. 6.10.2.1.1-1
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Dt
in ininw
=− ( )( ) = ≤
33 1 2 0 8550 580
54 1 150. .
.. [OK]
Check the flange proportions.
A Art. 6.10.2.2, A Eq. 6.10.2.2-1; A Eq. 6.10.2.2-2
bt
inin
f
f211 5
2 0 8556 73 12 0= ( )( ) = ≤.
.. . [OK]
b Df ≥
6
D in in6
33 1 2 0 8556
=− ( )( ). .
= 5.23 in ≤ bf = 11.5 in [OK]
tf ≥ 1.1 tw
A Eq. 6.10.2.2-3
1.1 tw = (1.1)(0.580 in) = 0.638
tf = 0.855 in ≥ 0.638 [OK]
Compare the moment of inertia of the compression flange of the steel sec-tion about the vertical axis in the plane of the web, Iyc, to the moment of inertia of the tension flange of the steel section about the vertical axis in the plane of the web.
0 1 10. ≤ ≤II
yc
yt
A Eq. 6.10.2.2-4
0.1 ≤ 1.0 ≤ 10 [OK]
Check the material thickness.
A Art. 6.7.3
tw = 0.580 in ≥ 0.25 in [OK]
137Design Examples
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Step 10: Check Live Load Deflection
A Art. 2.5.2.6.2; A Art. 3.6.1.3.2
The live load deflection should be taken as the larger of:
• That resulting from the design truck alone, or• That resulting from 25% of the design truck plus the design lane load
With Service I load combination, the maximum live load deflection limit is span/800.
The allowable service load deflection, which must be less than or equal to the span/800, is
A Art. 2.5.2.6.2
40 12
8000 6
ft inft in
( )
= .
For a straight multibeam bridge, the distribution factor for deflection, DFdeflection, is equal to the number of lanes divided by the number of beams.
A Comm. 2.5.2.6.2
DF no of lanesno of beamsdeflection =
= ..
24
= 0 5.
Deflection is governed by either design truck loading (HS-20) alone, or 25% of truck plus design lane load. See Figure 2.49.
A Art. 3.6.1.3.2
CL
6 ft
P1 = 32 kips P1 = 32 kips P2 = 8 kips
6 ft14 ft 14 ft
A B
FIGURE 2.49Position of design truck loading (HS-20) for deflection at midspan.
138 Simplified LRFD Bridge Design
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P DF P IMbeam deflection1 1 1
1000 5, .= ( )( ) +
= (( )( ) +( ) =32 1 0 33 21 3kips kips. .
P DF P IMbeam deflection2 2 1
1000 5, .= ( )( ) +
= (( )( ) +( ) =8 1 0 33 5 32kips kips. .
∆ ∆ ∆ ∆truck beam beam beamP P P= + +1 2 1, , ,
∆ ΣtruckPbxeEIL
L b x P LEI
= − −( )
+2 2 2 13
48
AISC Tbl. 3-23
= ( )( )( )( )
21 3 240 72
6 29 000 2
.
,
kips in inkipsin
( )( )
( ) − ( ) − (6710 480
480 408 724
2 2
in inin in in))( )
+ ( )( )( )( )
2
5 32 72 240
6 2
. kips in in
99 000 6710 480480
24
2
, kipsin
in inin
( )( )
( ) −− ( ) − ( )( )
+ ( )
72 408
21 3 48
2 2in in
kips. 00
48 29 000 6710
3
24
inkipsin
in
( )
( )
,
= 0.039 in + 0.010 in + 0.252 in = 0.301 in [controls]
Check requirements for 25% of truck and design lane load (0.64 kip/ft uni-formly distributed).
∆ lanewL
EI
kipsft
ftin
= =( )
5384
5 0 64 1124
.
( )
( )
480
384 29 000 6710
4
2
in
kipsin
i, nnin
40 189
( )= .
0.25Δtruck + Δlane = (0.25)(0.301 in) + 0.189 in = 0.264 in
The controlling deflection is
Δtruck = 0.301 in ≤ 0.6 in [OK]
139Design Examples
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Step 11: Check the Service Limit State
Permanent DeformationA Art. 6.10.4.2
Service II load combination should apply to calculate stresses in structural steel section alone.
A Art. 6.10.4.2.1
Flanges shall satisfy the requirements for steel flanges of noncomposite sections.
A Art. 6.10.4.2.2; Eq. 6.10.4.2.2-3
f f R Ffl
h yf+ ≤2
0 80.
where:ff = flange stress due to Service II loadsRh = hybrid factor = 1.0
A Art. 6.10.1.10.1Fyf = minimum yield strength of flangefl = flange lateral bending stress ≅ 0.0
The flange lateral bending stress, fl, is 0 because lateral bending stresses are assumed small.
0.80 RhFyf = (0.80)(1.0)(50 ksi)
= 40 ksi
The factored Service II moment for exterior beams is
Mu = 1.0(DC + DW) + 1.30(LL + IM)
= 1.0(322 ft-kips + 39.6 ft-kips) + 1.30(509.3 ft-kips)
= 1023.7 ft-kips (controls)
The factored Service II moment for interior beams is
Mu = 1.0 (16η ft-kips + 58.4 ft-kips) + 1.30 (545 ft-kips)
= 933.8 ft-kips
The section modulus, S, for a W33 × 130 is 406 in3.AISC Tbl.1-1
The flange stress due to the Service II loads calculated without consider-ation for flange bending is
f MS
f s inft
inf = =( )
1023 7 121
406 3
. t-kip
= 30.25 kips/in2 ≤ 0.80 RhFyf = 40 ksi [OK]
140 Simplified LRFD Bridge Design
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Step 12: Check Fatigue and Fracture Limit State
A Art. 6.6.1.2.2; A Art. 6.10.5.1
For load-induced fatigue considerations, each detail shall satisfy:
A Eq. 6.6.1.2.2-1
γ(Δf) ≤ (ΔF)n
where:γ = load factor for the fatigue load combination
= 0.75 for Fatigue II Limit State
A Tbl. 3.4.1-1
(Δf) = force effect, live load stress range due to the fatigue load
A Art. 3.6.1.4
(ΔF)n = nominal fatigue resistance (ksi)
A Art. 6.6.1.2.5
(ΔFTH) = constant amplitude fatigue threshold (ksi)
A Tbl. 6.6.1.2.5-3
For the Fatigue II load combination and finite life,
A Eq. 6.6.1.2.5-2
∆F ANn( ) =
1 3/
The constant taken for Detail Category A, Aout, is 250 × 108 kips/in2.
A Tbl. 6.6.1.2.5-1
The number of stress range cycles per truck passage, n, is 2.0 for simple span girders with span less than or equal to 40 ft.
A Tbl. 6.6.1.2.5-2
The number of design lanes, NL, is 2.0.
The constant amplitude threshold (ΔF)TH, for Detail Category A, is 24.0 kips/in2.
A Tbl. 6.6.1.2.5-3
12
12
24 0 122∆F kipsin
kips iTH( ) =
=. / nn2
141Design Examples
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The percentage of trucks (compared to all vehicles) in traffic for an urban interstate is 15%.
A Tbl. C3.6.1.4.2-1
This design example assumes that the average daily traffic, ADT, is 8,000 vehicles per day.
The average daily truck traffic is
ADTT = (0.15)(ADT) = (0.15)(8000 vehicles)(2 lanes) = 2400 trucks per day
The fraction of truck traffic in a single lane, p, is 0.85 when there are two lanes available to trucks.
A Tbl. 3.6.1.4.2-1
The single-lane average daily truck traffic, ADTTSL, is
ADTTSL = (p)ADTT = (0.85)(2400 trucks per day)A Eq. 3.6.1.4.2-1
= 2040 trucks per lane per day
The number of cycles of stress range for 75 years, N, isA Eq. 6.6.1.2.5-3
N = (365 days)(75 years)(n)AADTSL
where n = number of stress range cycles per truck passage = 2.0 for span lengths ≧ 40 ft
A Tbl. 6.6.1.2.5-2
N = (365 days)(75 years)(2.0 cycles per pass)(2040 trucks per day)
= 1.12 × 108 cycles
The nominal fatigue resistance, (ΔF)n, isA Eq. 6.6.1.2.5-2
∆F AN
x kipsin
x kipn( ) =
=1 3 8
2
8
250 10
1 12 10
/
. ssin
kips in2
26 07
= . /
Apply the fatigue limit state (excluding dead loads) using the maximum fatigue moment (exterior beam controls).
A Tbl. 3.4.1-1
142 Simplified LRFD Bridge Design
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Q = γMQ = γ[LL + IM]Q = ((0.75)(LL + IM)) = ((0.75)(229 ft-kips)) = 172 ft-kips
γ γ( )∆f MS
f ips inft
in= =
( )
172 121
406 3
t-k
γ(Δf) = 5.08 kips/in2 ≤ (ΔF)n = 6.07 kips/in2 [OK]
A Eq. 6.6.1.2.2-1
Special fatigue requirement for the web.A Art. 6.10.5.3
Vu ≤ Vcr
A Eq. 6.10.5.3-1
where:Vu = shear in the web due to the unfactored permanent load plus the fac-
tored fatigue loadVcr = shear-buckling resistance
A Eq. 6.10.9.3.3-1
Exterior girder governs in fatigue shear, thus use exterior girder distribution factor and shears.
The shear in the web at the section under consideration due to the unfac-tored permanent load plus the factored fatigue load is
Vu = VDC + VDW + Vfat
Vfat = γQ
= (0.75)(Vfatigue+IM)
= (0.75)(26.7 kips)
= 20.0 kips
Vu = 32.2. kips + 3.96 kips + 20.0 kips
= 56.16 kips
143Design Examples
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Nominal resistance of unstiffened webs isA Art. 6.10.9.2, 6.10.9.3.3
Vn = Vcr = CVp
A Eq. 6.10.9.3.3-1where:Vp = 0.58 FywDtw
= (0.58)(50 ksi)(31.39 in)(0.58 in) = 528 kipsC = ratio of the shear-buckling resistance to the shear yield strength.
A Art. 6.10.9.3.2
Find the ratio of shear buckling resistance to the shear yield strength, C, by satisfying the following equation:
If Dt
EkFw yw
≤ 1 12. , C = 1.0A Eq. 6.10.9.3.2-4
The shear-buckling coefficient k isA Eq. 6.10.9.3.2-7
kdD
o
= +
5 52
There are no specified transverse stiffeners, therefore do = infinity and k = 5.
Dt
in ininw
=− ( )( ) =
33 1 2 0 8550 580
54 1. .
..
1 12 1 1229 000 5
50
2. .
,EkF
kipsin
kipsi
yw=
( )
nn
Dtw
2
60 3 54 1= > =. . [OK]
Therefore, C = 1.0.The nominal shear resistance, Vn, of the web is
A Eq. 6.10.9.3.3-1Vn = Vcr = CVp = (1.0)(528 kips)
= 528 kips
Vu = 56.16 kips ≤ Vn = Vcr = 528 kips [OK]A Eq. 6.10.5.3-1
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Step 13: Check the Shear Adequacy at Support
Check that at the strength limit state the following requirement for shear is satisfied
Vu ≤ ΦvVn
A Eq. 6.10.9.1-1
The nominal shear resistance, Vn, is 528 kips.
Φv is 1.0 for steel structures.
For the Strength I Limit State, Vu, was determined in Step 8.
Vu = 156 kips [interior beam controls]
Vu = 156 kips ≤ ΦvVn = (1.0)(528 kips)
Design Example 5: Reinforced Concrete Slabs
Situation
The cast-in-place concrete deck for a simple span composite bridge is con-tinuous across five steel girders, as shown in Figure 2.50.
The overall width of the bridge is 48 ft.
The clear roadway width is 44 ft, 6 in.
The roadway is a concrete slab 9 in thick, with a concrete compres-sive strength at 28 days, f′c , of 4.5 kips/in2, and a specified minimum yield strength of the steel, Fy, of 60 kips/in2.
The steel girders are spaced at 10 ft as shown.
Allow for a 3 in future wearing surface, FWS, at 0.03 kips/ft2.
Assume the area of the curb and parapet on each side is 3.37 ft2.
The concrete self-weight, wc, is equal to 150 lbf/ft3.
Requirements
Design and review the reinforced concrete slab by the approximate method of analysis, AASHTO Art. 4.6.2.1. Use HL-93 loading.
A Arts. 4.6.2, 4.6.2.1
145Design Examples
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Solution
Step 1: Determine the Minimum Thickness of the Slab
The depth of a concrete deck should not be less than 7.0 in.A Art. 9.7.1.1
The assumed slab thickness is t = 8.5 in + 0.5 in for integral wearing sur-face. Use t = 9 in.
Step 2: Determine Dead Loads, w
The following dead loads are determined for a 1.0 ft wide transverse strip.
For a 9.0 in thick slab, including cantilever, the dead load is
w kipsft
in ftinslab =
( )
0 150 9 0 1123. . = 0 113 2. /kips ft
The dead load for the future wearing surface, wFWS, is given as 0.03 kips/ft2.The dead load for the curb and parapet in each side is
24 ft24 ft
10 ft 10 ft 10 ft 10 ft 4 ft4 ft
Clear width = 44 ft 6 in 1 ft 9 in1 ft 9 in
Curb andparapet
t
x = 0.66 ft
21 in
32 in–
4 ft
3 in9 in slab
W12 × 65
FIGURE 2.50Concrete deck slab design example.
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w kipsft
ft kipsC P& . . . /=
( ) =0 150 3 37 0 5063
2 fft of bridge span
Step 3: Find the Dead Load Force Effects
Approximate elastic method analysis as specified in Art. 4.6.2.1 is permitted.A Art. 9.6.1
The extreme positive moment in any deck panel between girders shall be taken to apply to all positive moment regions. Similarly, the extreme negative moment over any girder shall be taken to apply to all negative moment regions.
A Art. 4.6.2.1.1
The strips shall be treated as continuous beams with span lengths equal to the center-to-center distance between girders. For the purpose of determin-ing force effects in the strip, the supporting components (i.e., girders) shall be assumed to be infinitely rigid. The strips should be analyzed by classical beam theory.
A Art. 4.6.2.1.6
Apply loadings and determine the reaction at point A, as well as moments at points A, B, and C. These values will be representative of the maximum reaction, as well as the maximum positive and negative moments for the con-tinuous beam. Point A is located at the exterior support, point B at 0.4S inside the exterior support (4 ft to the right of point A), and point C at the first inte-rior support. It is noted that in the continuous beam, four equal spans and all spans loaded, the maximum positive moment in the first interior span occurs at 0.40S (AISC, 13th edition, continuous beam, four equal spans, all spans loaded). For clarification, these points are approximately located as shown in Figure 2.51.
For this analysis, the resultant reaction and moments can be calculated using the moment distribution method, influence line design aids, or a com-puter software program. In this example, a structural engineering software
S = 10 ft S = 10 ft S = 10 ft 4 ft4 ft
A B C
0.4S= 4 ft
FIGURE 2.51Locations in slab strips for maximum reactions and moments due to dead loads.
147Design Examples
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program was used to model and analyze the continuous slab. The loadings and results are summarized below.
The deck slab moments and reaction (excluding cantilever) are as follows and as shown in Figure 2.52.
MA = 0 kip-ft/ft
MB = +0.872 kip-ft/ft
MC = –1.211 kip-ft/ft
RA = 0.444 kips
The cantilever slab moments and reaction are as follows and as shown in Figure 2.53.
MA = –0.904 kip-ft/ft
MB = –0.439 kip-ft/ft
MC = +0.258 kip-ft/ft
RA = 0.568 kips
10 ft 10 ft
w = –0.113 kip/ft2
10 ft 4 ft4 ft 4 ft
AB
C
FIGURE 2.52Moments and reactions for deck slab dead load excluding deck cantilever.
10 ft 10 ft
w = –0.113 kip/ft2 w = –0.113 kip/ft2
10 ft 4 ft4 ft 4 ft
AB
C
FIGURE 2.53Moments and reaction for deck slab dead load in deck cantilever.
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The curb and parapet moments and reaction are as follows and as shown in Figure 2.54.
MA = –1.689 kip-ft/ft
MB = –0.821 kip-ft/ft
MC = +0.483 kip-ft/ft
RA = 0.723 kips
The future wearing surface moments and reaction are as follows and as shown in Figure 2.55.
MA = –0.076 kip-ft/ft
MB = +0.195 kip-ft/ft
MC = –0.300 kip-ft/ft
RA = 0.195 kips
10 ft 10 ft 10 ft 4 ft4 ft 4 ft
AB
0.66 ft0.66 ft
w = –0.506 kip/ft w = –0.506 kip/ft
C
FIGURE 2.54Moments and reaction for curb and parapet loads.
10 ft 10 ft 10 ft 4 ft4 ft 4 ft
1 ft 9 in 1 ft 9 in
A B C
w = –0.03 kip/ft2
FIGURE 2.55Moments and reaction for wearing surface loads.
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Step 4: Find the Live Load Force Effects
Where the approximate strip method is used to analyze decks with the slab primarily in the transverse direction, only the axles of the design truck shall be applied to the deck slab.
A Art. 3.6.1.3.1; A Art. 3.6.1.3.3
Wheel loads on the axle are equal and transversely spaced 6.0 ft apart.A Fig. 3.6.1.2.2-1
The design truck shall be positioned transversely to find extreme force effects such that the center of any wheel load is not closer than 1.0 ft from the face of the curb for the design of the deck overhang and 2.0 ft from the edge of the design lane for the design of all other components.
A Art. 3.6.1.3.1
Find the distance from the wheel load to the point of support, X, where S is the spacing of supporting components using Figure 2.56.
For this example, force effects are calculated conservatively using concen-trated wheel loads.
A Art. 4.6.2.1.6
Generally the number of design lanes to be considered across a transverse strip, NL, should be determined by taking the integer part of the ratio w/12, where w is the clear roadway width in ft and 12 ft is the width of the design lane.
A Art. 3.6.1.1.1
21 in. 12 in.
P = wheel load P = wheel load
48 in.
3 in. future wearing surface
6 ft
15 in. = X
9 in. slab
A
FIGURE 2.56Live load placement for maximum negative moment.
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For this example the number of design lanes is
N ftft
lane
lanesL = =44 5
123 7. . (3 lanes)
The multiple presence factor, m, is 1.2 for one loaded lane, 1.0 for two loaded lanes, and 0.85 for three loaded lanes. (Entries of greater than 1 in [A Tbl. 3.6.1.1.2-1] result from statistical calibration on the basis of pairs of vehicles instead of a single vehicle.)
A Art. 3.6.1.1.2; Tbl. 3.6.1.1.2-1
At this point, the live loads are applied to the deck to find once again the resulting reaction at point A and moments at points A, B, and C.
Find the maximum negative live load moment for overhang.
The critical placement of a single wheel load is at X = 1.25 ft. See Figure 2.57.The equivalent width of a transverse strip for the overhang is
A Tbl. 4.6.2.1.3-1
45 0 10 0 45 0 10 0 15
1257 5. . . . .+ = + ( )
=X inin
in == 4 79. ft
The multiple presence factor, m, is 1.2 for one loaded lane causing the max-imum moment.
Therefore, the negative live load moment for overhang is
Mkips ft
ftA =−( )( )( ) = −
1 2 16 0 1 254 79
5 01. . .
.. ft-kips/ft width
Find the maximum positive live load moment in the first interior span.
For internalmoment at point A,cut at point A
A
MA
P = 16 kips
X = 15 in= 1.25 ft
FIGURE 2.57Live load placement for maximum negative moment, one lane loaded.
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For a continuous concrete slab with repeating equal spans with all spans loaded with dead loads, the largest positive bending moment occurs at point B, which is 0.4S or 4.0 ft from point A. For this analy-sis, both single and double lane loadings need to be investigated.
For both cases, the positive moment in the equivalent strip width is
26.0 + 6.6S = 26.0 + (6.6)(10 ft) = 92 in = 7.67 ftA Tbl. 4.6.2.1.3-1
To start, a wheel load is located at point B (0.4S), with the other wheel load 6 ft away as dictated by the design truck in the H-93 load model (See AASHTO Art. 3.6.1.2.2). Please see Figure 2.58.
Using structural analysis software or influence line diagrams,
RA = 8.160 kips
MB = 32.64 ft-kips
The previous values must be further adjusted to account for strip width and the multiple presence factor, m, of 1.2. The maximum positive live load moment in the first interior span is,
A Art. 3.6.1.1.2
′ = ( )( ) = ( )( ) =Rm R
ftkips
ftAAA
7 671 2 8 16
7 671
.. .
..228 kips per foot width
′ = ( )( ) = ( )( )Mm M
ftf ipsfBB
B
7 671 2 32 64
7 67.. .
.t-ktt
= 5 11. ft-kips per foot width
10 ft10 ft 10 ft 10 ft 4 ft4 ft
0.4S = 4 ft
P = 16 kips P = 16 kips
6 ft
A
B
C
FIGURE 2.58Live load placement for maximum positive moment in first interior span, one lane loaded.
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If the loadings were placed in the second interior span, they would decrease the effects of the first truck loading in the first interior span.
For double lane loadings, the multiple presence factor, m, is 1.0 at points B and C. Another set of wheel loads is added in the third span, as shown at X = 28 ft and 34 ft from the slab end.
This loading will cause the first interior span moment to increase at B.For two lanes loaded, see Figure 2.59.
By a structural analysis software program,
RA = 8.503 kips
MB = 34.01 ft-kips
Modifying the previous values for strip width and two lanes loaded,
′ = ( )( ) = ( )( ) =Rm R
ftkips
ftSA
7 671 0 8 503
7 671
.. .
..111 kips per foot width
′ = ( )( ) = ( )( )Mm M
ftf ipsfBB
B
7 671 0 34 01
7 67.. .
.t-ktt
= 4 43. ft-kips per foot width
Thus, the single lane loaded case governs, and the moment effect decreases in the first interior span with increased lane loading cases. Therefore, a sce-nario of three loaded lanes does not need to be considered for the maximum positive live load moment in the first interior span.
Find the maximum negative live load moment in the first interior span.
10 ft10 ft 10 ft 10 ft 4 ft4 ft
x
A
B
C
4 ft
P =16 kips
P =16 kips
6 ft 4 ft
P =16 kips
P =16 kips
6 ft
FIGURE 2.59Live load placement for maximum positive moment, double lane loaded.
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The critical location for placement of the design truck load for maximum negative moment occurs at the first interior deck support under a one lane load case. Please see Figure 2.60.
The equivalent transverse strip width isA Tbl. 4.6.2.1.3-1
48.0 + 3.0S = 48.0 + (3.0)(10 ft) = 78 in = 6.5 ft
Using a structural analysis software program, the maximum negative live load moment at first interior span is,
MC = –27.48 ft-kips
m = 1.2 for one loaded lane
′ = ( )( ) = ( ) −( )Mm M
ftips
ftCCC
6 51 2 27 48
6 5.. .
.ft-k
== −5 07. ft-kips per foot width
The small increase due to the loading of the second truck is not enough to consider because m = 1.0 for two loaded lanes. Therefore, it is only necessary to consider the one lane loaded case.
Find the maximum live load reaction at point A.
The exterior wheel load is placed 1.0 ft from the curb. The width of the strip is conservatively taken as the same as for the overhang. The governing load-ing is shown in Figure 2.61.
A Art. 3.6.1.3.1
10 ft10 ft 10 ft
7 ft
3 ft3 ft
10 ft 4 ft4 ft
x
P =16 kips
P =16 kips
A
B
C
7 ft
FIGURE 2.60Live load placement for maximum negative moment in first interior span, one lane loaded.
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The equivalent strip width of a cast-in-place deck for overhang, where x is the distance from the load to the point of support in feet, is
A Tbl. 4.6.2.1.3-1
45.0 + 10.0x = 45.0 + 10.0(1.25 ft) = 57.5 in = 4.79 ft
Using a structural analysis software program, when the exterior wheel load is 1.0 ft from the centerline of support, RA is 24.42 kips. When the exte-rior wheel load is 1.0 ft from the curb, RA is 25.36 kips (controls). Please see Table 2.13.
′ = = ( )( )R mR kipsAA
A
equiv strip width1 2 25 36
4 7. .
. 996 35
ft= . kips per foot width
10 ft10 ft 10 ft
6 ft
10 ft 4 ft4 ft
21 in + 12 in = 33 in
15 in
x
P =16 kips
P =16 kips
A
B
C
FIGURE 2.61Live load placement for maximum reaction at first support.
TABLE 2.13
Force Effects Summary Table
RA
(kips/ft)MA
(ft-kips/ft)MB
(ft-kips/ft)MC
(ft-kips/ft)
Deck slab excluding cantilever 0.444 0.0 0.872 –1.211Cantilevered slab (overhang) 0.568 –0.904 –0.439 0.258Curb and parapet 0.723 –1.689 –0.821 0.483Future wearing surface, FWS 0.195 –0.076 0.195 –0.300Max negative moment due to overhang — –5.010 — —Max LL reaction and moment (first span) 1.280 — 5.110 —Max negative LL moment (first span) — — — –5.070Max LL reaction at first support A 6.350 — — —
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Step 5: Perform the Strength I Limit State Analysis
Each deck component and connection shall satisfy the following equation.A Art. 1.3.2.1
For loads for which a maximum value of load factor, γi, is appropriate,A Eq. 1.3.2.1-2
ηi = ηDηRηI ≥ 0.95
For loads for which a minimum value of load factor, γi, is appropriate,A Eq. 1.3.2.1-3
ηη η ηi
D R I= ≤1 1 0.
ηD = 1.0A Art. 1.3.3
ηR = 1.0A Art. 1.3.4
ηI = 1.0A Art. 1.3.5
The load factor for permanent loads, γp, is taken at its maximum value if the force effects are additive and at the minimum value if it subtracts from the dominant force effects. See Table 2.14.
A Tbl. 3.4.1-2
The dynamic load allowance, IM, is 33% of the live-load force effect. A Art. 3.6.2.1
For the Strength I Limit state, the reaction and moments are,
Q = ΣηiγiQi = (1.00)γpDC + (1.00)γpDW + (1.00)(1.75)(LL + IM)A Tbl. 3.4.1-1, Tbl. 3.4.1-2
TABLE 2.14
Load Factors for Permanent Loads
γp Max Min
DC 1.25 0.9DW 1.5 0.65
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DW includes the future wearing surface load only, and DC represents all other permanent loads.
RA = (1.0)(1.25)(0.444 kips/ft + 0.568 kips/ft + 0.723 kips/ft)
+ (1.0)(1.5)(0.195 kips/ft)
+ (1.0)(1.75)(6.35 kips/ft + (0.33)(6.35 kips/ft))
= 17.24 kips per foot width
MA = (1.0)(1.25)(–0.904 ft-kips/ft + (–1.689 ft-kips/ft) + (1.0)(1.5)(–0.076 ft-kips/ft)
+ (1.0)(1.75)(–5.010 ft-kips/ft + (0.33)(–5.010 ft-kips/ft))
= –15.02 ft-kips per foot width
MB = (1.0)(1.25)(0.872 ft-kips/ft + (1.0)(0.9)(–0.439 ft-kips/ft + (–0.820 ft-kips/ft)))
+ (1.0)(1.5)(0.195 ft-kips/ft)
+ (1.0)(1.75)(5.110 ft-kips/ft + (0.33)(5.110 ft-kips/ft))
= 12.14 ft-kips per foot width
MC = (1.0)(1.25)(–1.211 ft-kips/ft + (1.0)(0.9)(0.258 ft-kips/ft + 0.483 ft-kips/ft))
+ (1.0)(1.5)(–0.300 ft-kips/ft)
+ (1.0)(1.75)(–5.070 ft-kips/ft + (0.33)(–5.070 ft-kips/ft))
= –13.10 ft-kips per foot width
It is noted that the load factor for cantilevered slab and curb/parapet load, γp, is 0.90 for producing maximum value for MB and MC.
A Tbl. 3.4.1-2
For the selection of reinforcement, the negative moments may be reduced to their value at a quarter flange width from the centerline of support. For the purposes of simplicity, this calculation is not performed in this example, but can easily be achieved using classical methods of structural analysis. See Table 2.15.
A Art. 4.6.2.1.6
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Step 6: Design for Moment
The compressive strength of the concrete at 28 days, f′c , is 4.5 kips/in2. The specified minimum yield point of the steel, Fy, is 60 kips/in2. Determine the configuration of epoxy-coated steel reinforcement required.
A Art. 5.10.3.2
The maximum spacing of primary reinforcement for slabs is 1.5 times the thickness of the member or 18.0 in. By using the structural slab thickness of 8.5 in, the maximum spacing of reinforcement becomes:
smax = (1.5)(8.5 in) = 12.75 in < 18 in [controls]
The required concrete cover is 2.5 in for the unprotected main reinforce-ment for deck surfaces subject to wear and 1.0 in for the bottom of cast-in-place slabs. For conservatism and simplicity of this example, 2.5 in is used for both top and bottom covers.
A Tbl. 5.12.3-1
Note that if epoxy-coated bars are used, 1.0 in is required for covers for top and bottom.
A Art. 5.12.3Assuming no. 5 steel rebar,
db = 0.625 in
Ab = 0.31 in2
See Figure 2.62 for the reinforcement placement.
Factored moment MA = –15.02 ft-kips is equal to the factored flexural resistance Mr.
The distance from the extreme compression fiber to the centroid of reinforcing bars is
d = 9.0 in – 0.5 in (integral wearing surface) – 2.5 in (cover) – 0.625 in/2 (bar diameter)
= 5.68 in
TABLE 2.15
Strength I Limit State Summary (Factored Values)
Maximum extreme positive moment, MB 12.14 ft-kips/ftMaximum extreme negative moment, MA –15.02 ft-kips/ft > Mc = –13.10 ft-kips/ftMaximum reaction, RA 17.24 kips/ft
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The depth of the equivalent stress block is
aA f
f bA ksi
ksi ins y
c
s=′
= ( )( )( )(0 85
600 85 4 5 12. . . )) = 1 307. As
A Art. 5.7.3.2The factored resistance Mr is
M M A f d ar n s y= = −
Φ Φ2
where:Φ = resistance factor = 0.9 for flexure in reinforced concreteMn = nominal resistance
Let Mr equal to the factored moment Mu.
Mu (= MA) = Mr
15 02 12 0 9 60. .f ips
ftinft
A ksist-k
= (( ) −
5 68 1 3072
. .in As
A Art. 5.5.4.2.1The minimum area of steel needed is
As = 0.64 in2/ft of slab width
Use no. 5 bars spaced at 5.5 in. So, the provided area of steel is
A A in widthspacing
in inis b=
=12 0 31 125 5
2.. nn
inft
= 0 682
. > As = 0.64 in2/ft [good]
Check the moment capacity,
aA f
f bin ksi
ksis y
c=
′=
( )( )( )0 85
0 68 600 85 4 5
2
..
. .(( )( ) =12
0 889in
in.
12 in
d
2.5 in#5 steel rebar
9.0 int =
FIGURE 2.62Deck slab section for reinforcement placement.
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Confirm that Mr is equal to or greater than the factored moment MA (= Mu)
M M A f d ar n s y= = −
ϕ ϕ2
= ( )( )( ) −
0 9 0 68 60 5 68 0 8892
2. . . .in ksi in in 1112
ftin
= 16.0 ft-kips/ft > Mu = 15.02 ft-kips/ft [OK]
Check minimum steel.
The minimum reinforcement for flexural components is satisfied if a fac-tored flexural resistance ΦMn = Mr is at least equal to the lesser of 1.2 times the cracking moment, Mcr, and 1.33 times the factored moment required by the applicable strength load combination.
A Art. 5.7.3.3.2
Where slabs are designed for a noncomposite section to resist all loads, the cracking moment is,
A Eq. 5.7.3.3.2-1
Mcr = Sncfr
where,Snc = section modulus of the noncomposite sectionfr = modulus of rupture
For normal weight concrete, the modulus of rupture of concrete is
A Art. 5.4.2.6
f f kips in kips inr c= ′ = =0 37 0 37 4 5 0 7852 2. . . / . /
The section modulus for the extreme fiber of the noncomposite section where tensile stress is caused by external loads is
S in in innc =
( )( ) =1
612 8 5 144 52 3. .
The cracking moment is
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M S f in kipsincr nc r= = ( )
=144 5 0 785 11332. . .44 i psn-ki
1 2 1 2 113 4 112
. . .M i ps ftincr = ( )( )
n-ki
= 11.34 ft-kips/ft
Mu = MA
1 33 1 33 15 02 19 98. . . .M f psftu = ( )
=t-ki ft-kiips ft/
1.2 Mcr = 11.34 ft-kips/ft [controls]
11.34 ft-kips/ft < ΦMn = Mr = 16.0 ft-kips/ft [OK]
Reinforcement transverse to the main steel reinforcement (which is per-pendicular to traffic) is placed in the bottom of all slabs. The amount shall be a percentage of the main reinforcement required as determined in the following formula.
A Art. 9.7.3.2
For primary reinforcement perpendicular to traffic, Se is the effective span length; the following must be true.
A Art. 9.7.3.2
220 67
Se≤ %
For slabs supported on steel girders, Se is the distance between flange tips, plus the flange overhang, taken as the distance from the extreme flange tip to the face of the web, disregarding any fillet. (For a W12 × 65, tw = 0.39 in).
A Art. 9.7.2.3
S ft in ftin
fte = − ( )
=10 0 39 112
9 97. .
Se ≅ 10 ft
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Reinforcement shall be placed in the secondary direction in the bottom of slab as a percentage of the primary reinforcement perpendicular to traffic for positive moment as follows:
A Art. 9.7.3.2
220 67Se
≤ %
220 22010
69 6 67S fte
= = > ( ). % % [no good]
Therefore, use 67%.
As = 0.67 As = (0.67)(0.68 in2) = 0.46 in2/ft
For longitudinal bottom bars, use no. 5 (Ab = 0.31 in2) at 8 in,
A A in width
spacingin in
is b=
= ( )12 0 31 128
2.nn
= 0.46 in2/ft
The reinforcement needed in each direction for the shrinkage and tem-perature reinforcement shall be
A Art. 5.10.8
A bhb h fs temp
y,
.≥+( )
1 302
where:b = least width of component section (12 in)h = least thickness of component section (8.5 in)fy = specified yield strength of reinforcing bars ≤ 75 kips/in2
A Eq. 5.10.8-1
Ain in
in in kips temp,. .
.≥ ( )( )
+( )1 30 12 8 5
2 12 8 5 60 ssin
inft
2
20 054
= .
0.11 ≤ As,temp ≤ 0.60, so use #3 bars (Ab = 0.11 in2/ft)A Eq. 5.10.8-2
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The primary and secondary reinforcement already selected provide more than this amount; however, for members greater than 6.0 in thickness the shrinkage and temperature reinforcement is to be distributed equally on both faces. The maximum spacing of this reinforcement is 3.0 times the slab thickness or 18 in. For the top face longitudinal bars, the area of the tempera-ture reinforcement, As,temp, is 0.11 in2/ft. Use no. 4 bars at 18 in, providing As = 0.13 in2/ft.
A Eq. 5.10.8
For primary reinforcement, use no. 5 bars at 5.5 in. For longitudinal bottom bars, use no. 5 at 8 in. For longitudinal top bars use no. 4 at 18 in.
Step 7: Alternate Solution Utilizing Empirical Design Method
The empirical method is a simplified design method that can be utilized when provisions of Art. 9.7.2 are satisfied. This method does not require extensive structural analysis (by the moment distribution method, influence line aids, or computer methods) as required in the previously worked solution. As such, the empirical method may prove useful in written testing situations.
Empirical design shall not be applied to deck overhangs.
A Art. 9.7.2.2
Verify that requirements for the empirical design method use are satisfied.
A Art. 9.7.2.4
Find the effective length of the slab cut out or:
A Art. 9.7.2.3
For a W 12 × 65 steel girder, the web thickness, tw, is 0.39 in.
L ft t ft in ftineffective w= − = − ( )
10 10 0 39 112
.
= 9 97. ft
Effective length of slab cannot exceed 13.5 ft. [OK]
A Art. 9.7.2.4
Ratio of effective length to slab depth cannot exceed 18.0 and is not less than 6.0.
A Art. 9.7.2.4
The slab depth, d, is
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d in ftin
ft= ( )
=9 112
0 75.
L
dftft
effective = =9 940 75
13 3..
. [OK]
Slab thickness is not less than 7.0 in excluding wearing surface. [OK]A Art. 9.7.2.4
Overhang beyond center line of outside girder at least equals 5.0 times depth of slab.
A Art. 9.7.2.4
4 ft > (5.0)(0.75 ft) = 3.75 ft [OK]
Slab core depth is not less than 4.0 in. [OK]A Art. 9.7.2.4
The following conditions are also met or assumed to be satisfied:A Art. 9.7.2.4
• Cross frames or diaphragms are used throughout.• Deck is uniform depth, excluding haunches.• Supporting components are made of steel.• Concrete strength, f′c , is at least 4000 psi.• Deck is made composite with supporting structural components
(in both positive and negative moment regions, with adequate shear connectors).
Four layers of isotropic reinforcement are required.A Art. 9.7.2.5
Reinforcement shall be closest to outside surface as possible.A Art. 9.7.2.5
Outermost reinforcing layer shall be in effective direction.A Art. 9.7.2.5
Spacing shall not exceed 18.0 in.A Art. 9.7.2.5
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Reinforcing shall be Grade 60 steel or better.A Art. 9.7.2.5
Design top reinforcing steel:
For top layer, 0.18 in2/ft is required in each direction.A Art. 9.7.2.5
Try #4 bars with 12 in spacing.Total area of steel is equal to 0.198 in2.At 12 in spacing, we have 0.198 in2 per foot.For bottom layer, 0.27 in2/ft is required in each direction.
A Art. 9.7.2.5
Try #5 bars spaced at 12 in.Total area of steel = 0.309 in2.At 12 in spacing, we have 0.309 in2.For bottom reinforcing, use #5 spaced at 12 in in each direction.
It can be seen that significantly less reinforcing bar is required when designed by the empirical method. This is due to a complex arching affect that governs behavior of the slab in these conditions.
A Art. C9.7.2.1
Design Example 6: Prestressed Interior Concrete Girder
Situation
An interior prestressed concrete girder for a two-lane simply supported highway bridge in central New York State is to be designed. The spacing of the bridge’s five girders is 7 ft 6 in. The width of the exterior beam overhang is 3 ft 9 in.
The design load is AASHTO HL-93. Allow for a future wearing surface, FWS, of 3 in bituminous concrete with a load, wFWS, of 0.140 kips/ft3, and use Strength I load combination for load resistance factor design.
L bridge span 80 ft integral wearing surface of slab 0.5 inWFWS load of future wearing surface of 3 in bituminous pavement 0.140 kips/ft3
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Aps area of grade 270 prestressing steel (44 strands at ½ in diameter; 7 wire = (44)(0.153 in2)) 6.732 in2
Es modulus of elasticity of prestressing steel 28500 ksif′cg = f′c compressive strength of concrete at 28 days for prestressed I-beams 6500 lbf/in2
f′ci compressive strength of concrete at time of initial prestress 6000 lbf/in2
f′cs compressive strength of concrete for 8 in slab 4500 lbf/in2
fpu specified tensile strength of prestressing steel 270 kips/in2
The basic beam properties are as follows:
Ag cross-sectional area of basic beam 762 in2
Ig moment of inertia of basic beam about centroidal axis, neglecting reinforcement 212,450 in4
Snc,top top section modulus for the extreme fiber of the noncomposite section 7692 in3
Snc,bottom bottom section modulus for the extreme fiber of the noncomposite section 9087 in3
yb distance from the bottom fiber to the centroid of the basic beam 23.38 in
Please see Figure 2.63.
80 ft
(a) Elevation
1.75 ft
3.33 ftt = 8 in
38 ft(b) Cross section
4 ft7.5 ft
51in 24 in
7.5 ft7.5 ft7.5 ft4 ft
FIGURE 2.63Prestressed concrete interior girder design example.
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Requirements
Determine the composite section properties; the factored design moment for Strength I Limit State at midspan, Mu; and the girder moment capacity, ΦMn (= Mr). For load combination Limit State Service I, determine the concrete stresses in the girder at midspan at release of prestress and the final concrete stresses after all losses (except friction) at midspan. Please see Figure 2.64.
Step 1: Determine the Composite Section Properties
The thickness of the following sections of precast concrete beams must meet the following specifications:
A Art. 5.14.1.2.2
top flange > 2 in [OK]web > 5 in [OK]bottom flange > 5.0 in [OK]
0.5 in integralwearing thickness
90 in
7.5 in
11 in18 in
17 in
10 in
9 in
c.g. basic beam
em = 23.38 in – 5 in= 18.38 in
8 in 8 in 4 spacesat 2 in = 8 in
51 in
yt =27.62 in
yb =23.38 in
12 spacesat 2 in = 24 in
44 @ ½ in diameterstrands (seven wire)
4 in
5 in
Centroid of prestressingstrands pattern = g = 5 in
#
FIGURE 2.64Cross section of girder with composite deck.
167Design Examples
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Calculate the minimum depth including deck for precast prestressed con-crete I-beams with simple spans, dmin.
A Tbl. 2.5.2.6.3-1
dmin = 0.045 L = (0.045)(80 ft)(12 in/ft) = 43.2 in
d = 7.5 in + 51 in = 58.5 in > 43.2 in [OK]
btop top flange width 18 inL effective span length (actual span length) 80 ftts slab thickness 7.5 in
The effective slab flange width for interior beams isA Art. 4.6.2.6.1
bi = be = S = (7.5 ft)(12 in/ft) = 90 in
The normal unit weight of concrete, wc, is 0.15 kips/ft3.A Tbl. 3.5.1-1
The modulus of elasticity of concrete at 28 days for prestressed I-beams is
A Eq. 5.4.2.4-1
E w fcg c cg= ′33 000 1 5, .
= ( )
33 000 0 15 6 53
1 5
2, . ..kips
ftkipsin
= 4890 kips/in2
The modulus of elasticity of the slab is
E w fs c cs= ′33 000 1 5, .
= ( )
33 000 0 15 4 53
1 5
2, . ..kips
ftkipsin
= 4070 kips/in2
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The modular ratio is
nEE
kipsinkipsin
cg
cs= = =
4890
40701 2
2
2
.
The transformed area of the slab is
7 5 901 2
562 5 2..
.in in in( )
=
Please see Figure 2.65.
The area of the composite section, Ac, is
Ac = Ag + transformed area of slab
= 762 in2 + 562.5 in2
= 1324.5 in2
90 in/1.2 = 75 in
7.5 in + 0.5 in
11 in 18 in
17 in
10 in
9 in
51 in58.5 in
24 in
c.g. basicbeamsection
c.g. ofcompositesection
y b = yb + y
y t = 21.8 in
= 36.7 in
–
y =13.32 in–
yt =27.62 in
yb =23.38 in
4 in
Centroid of prestressingstrand pattern = g = 5 in
FIGURE 2.65Area transformed section of girder section.
169Design Examples
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From the centroid of the basic beam section, c.g., solve for y.
1324 5 562 5 7 52
2 2. . .in y in y int( ) = ( ) +
yin in in
in=
( ) +( )562 5 27 62 3 751324 5
2
2
. . ..
′ = + = + =y y y in in inbb b 23 38 13 32 36 7. . .
′ = +( ) − ′ =y in in y intt bb51 7 5 21 8. .
The composite moment of inertia, Ic, is
I I A yin in
in yc g g tt= + + ( )( ) + ( ) ′ −23
275 7 512
562 5.
. 77 52
2. in
= 212,450 in4 + (762 in2)(13.32 in)2 + 2636.7 in4 + (562.5 in2)(21.8 in – 3.75 in)2
= 533,546.5 in4
For the bottom extreme fiber, the composite section modulus is
S Iy
inin
inbcc
bb=
′= =533 546 5
36 714 538 0
43, .
., .
For the top extreme fiber (slab top), the composite section modulus is (see Table 2.16),
S Iy
inin
intcc
tt=
′= =533 546 5
21 824 474 6
43, .
., .
TABLE 2.16
Summary of Section Properties
Basic Composite
Ag 762 in2 Ac 1324.5 in2
yt 27.62 in y′t 21.8 inyb 23.38 in y′b 36.7 inIg 212,450 in4 Ic 533,546.5 in4
Snct 7692 in3 Stc 24,474.6 in3
Sncb 9087 in3 Sbc 14,538 in3
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Step 2: Determine the Factored Maximum Moment
Use AASHTO HL-93 live load model to find the unfactored moment and shear due to live load.
For all components excluding deck joints and fatigue, the dynamic load allowance, IM, is 33%.
A Art. 3.6.2.1The beam spacing, S is 7.5 ft.The bridge span, L, is 80 ft.
Calculate the distribution factor for moments, DFMsi, for interior beams with one lane loaded.
A Tbl. 4.6.2.2.2b-1 or Appendix A
NOT E: The multiple presence factor, m, is already included in the approxi-mate equations for live load distribution factors.
A Art. C3.6.1.1.2
KLtg
s121 093
0 1
=.
. for the cross-section type (k).
A Tbls. 4.6.2.2.1-1; 4.6.2.2.1-2; 4.6.2.2.2b-1 or Appendix A
DFM S SL
KLtsig
s= +
0 06
14 12
0 4 0 3
3.. .
0 1.
= +
0 06 7 514
7 580
1 00 4 0 3
. . . .. .
ft ftft
99 0 477( )
= .
Calculate the distribution factor for moments, DFMmi, for interior beams with two lanes loaded.
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
= +
0 075 7 59 5
7 580
10 6 0 2
. ..
.. .ft ft
ft.. .09 0 664( )
= [controls]
171Design Examples
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Approximate maximum bending moments at midspan due to HL-93 load-ing as follows and shown in Figure 2.66.
Find the truck load (HS-20) moment, Mtr
ΣMB = 0
0 = 8 kips(26 ft) + 32 kips(40 ft) + 32 kips(54 ft) – RA(80 ft)
RA = 40.2 kips
Mtr = (40.2 kips)(40 ft) – 32 kips(14 ft)
= 1160 ft-kips
Find the tandem load moment Mtandem
ΣMB = 0
0 = 25 kips(36 ft) + 25 kips(40 ft) – RA(80 ft)
80 ft
(a) Design truck load (HS20) position
(b) Design tandem load position
w = 0.64 kips/ft
4 ft 36 ft40 ft
14 ft 14 ft 26 ft26 ft
32 kips 32 kips
25 kips25 kips
8 kips
A B
A B
(c) Design lane load position
A B
FIGURE 2.66Bending moments at midspan due to HL-93 loading.
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RA = 23.75 kips
Mtandem = (23.75 kips)(40 ft) = 950 ft-kips
Find the lane load moment, Mln
M
kipsft
kipsft fln
.=
=
0 64 80
8512
2
t-kkips
The maximum live load plus impact moment per girder is defined by the following equation:
M DFM M or M IM MLL IM mi tr+ = ( ) +
+tandem 1100 ln
= (0.664)((1160 ft-kips)(1 + 0.33) + 512 ft-kips)
= 1364.38 ft-kips per girder
Find the moment due to DC, MDC. The beam weight is
w in ftin in
kipsft
= ( ) ( )( )
762 1
12 120 152
2
3.
= 0.79 kips/ft
The moment due to the beam weight is
M wLkips
ftft
fg =
=
( )
=2
2
8
0 79 80
8632
.tt-kips
The slab dead load is
w in ftin
kipsft
f= ( ) ( )
8 112
0 15 7 53. . tt( )
= 0.75 kips/ft
173Design Examples
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The moment due to the slab dead load is
M wLkips
ftft
fD =
=
( )
=2
2
8
0 75 80
8600
.tt-kips
The total moment due to DC is
MDC = Mg + MD = 632 ft-kips + 600 ft-kips = 1232 ft-kips
The future wearing surface is 3 in.Find the moment due to DW, MDW.
w ininft
kipsftDW =
3 0
120 140 73
. . .55 0 263ft kips ft( ) = . /
The moment due to DW is (also see Table 2.17)
M w Lkips
ftft
DWDW=
=
( )
=2
2
8
0 263 80
8
.2210 4. ft-kips
Limit StatesA Art. 1.3.2
Find load modifier ηi.
ηi = ηDηRηI ≥ 0.95
ηD ductility factor 1.00 for conventional designA Art. 1.3.3
TABLE 2.17
Unfactored Moments per Girder
Load Type Moment (ft-kips)
DC 1232.0DW 210.4LL + IM 1364.38
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ηR redundancy factor 1.00 for conventional levels of redundancy
A Art. 1.3.4
ηI operational importance factor 1.00 for typical bridges
A Art. 1.3.5
Find the factored maximum moment for the Strength I Limit State
A Tbls. 3.4.1-1; 3.4.1-2
Q = Σηi γi Qi
Q = 1.0[γp DC γp DW + 1.75 (LL + IM)]
Mu = (1.0)(1.25)(1232.0 ft-kips) + (1.50)(210.4 ft-kips) + (1.75)(1364.38 ft-kips)
= 4243.26 ft-kips
Step 3: Determine the Girder Moment Capacity, ΦMn (= Mr)
Find the average stress in prestressing steel.Assume rectangular behavior.Find the distance from the extreme compression fiber to the neutral axis, c.
Aps area of the prestressing steel 6.732 in2
b width of the compression face member 90.0 in
dp distance from extreme compression fiber to the
centroid of the prestressing tendons in
f′c compressive strength of concrete at 28 days 6.5 kips/in2
fps average stress in prestressing steel at the time for
which the nominal resistance of member is required kips/in2
fpu specified tensile strength of prestressing steel 270 kips/in2
fs stress in the tension reinforcement 0 kips/in2
f′s stress in the compression reinforcement 0 kips/in2
k shear-buckling coefficient for webs 0.28
A Tbl. C5.7.3.1.1
ts slab thickness 8.0 in
175Design Examples
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The distance from the compression fiber to the centroid of prestressing tendons is
dp = (51 in – 5 in) + 8 in = 54 in
The factor for concrete strength isA Art. 5.7.2.2
β1 = 0.85 – (0.05)(f′c – 4 ksi) = (0.05)(6.5 ksi – 4 ksi)
= 0.725
The distance from the extreme compression fiber to the neutral axis is
A Eq. 5.7.3.1.1-4
cA f A f F f
f b kAfd
ps pu s s s s
c psps
p
=+ − ′ ′
′ +0 85 1. β
=
( )( ) + ( )( ) − ( )6 732 270 0 0 0 02 2. in ksi in ksi ksi ksi(( )( )( )( )( ) + ( )0 85 6 5 0 725 90 0 28 6 732. . . . .ksi in inn ksi
in2 270
54 0( ) .
= 4.91 in < ts = 8 in [assumption OK]
The average stress in the prestressing steel isA Eq. 5.7.3.1.1-1
f f k cdps pu
p= −
1
= ( ) − ( )
=270 1 0 28 4 91
54263ksi in
in. . ..1 ksi
Find the flanged section factored flexural resistance
A Art. 5.7.3.2.2
a = β1c = (0.725)(4.91 in) = 3.56 in
a depth of equivalent rectangular stress block 3.56 in
Φ resistance factor 1.00
A Art. 5.5.4.2
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The nominal flexural resistance, Mn, and the factored resistance, Mr, are,A Art. 5.7.3.2
Neglecting nonprestressed reinforcement,
M A f d an ps ps p= −
2
M in ksi in inn = ( )( ) −
=6 732 263 7 54 3 562
72. . . 7725 20. ft-kips
Mr = ΦMn = (1.0)(7725.20 ft-kips) > Mu
= 4243.46 ft-kips for Strength I Limit State [OK]
Step 4: Determine Concrete Stresses at Midspan at Release of Prestress
Temporary allowable concrete stresses before losses (at time of initial pre-stress) due to creep and shrinkage are as follows. See Figure 2.67.
51 in
18 in
24 in
c.g. of the basicgirder section
yb = 23.38 in
g = 5 in
= em = averageprestressingsteeleccentricityat midspan= 18.38 in
FIGURE 2.67Girder I-beam section.
177Design Examples
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For compression,
A Art. 5.9.4.1.1 fci = 0.6 f′ci = (0.6)(6.0 ksi) = 3.6 ksi
For tension,
A Art. 5.9.4.1.2
f f ksi ksiti ci= ′ = =0 24 0 24 5 5 0 563. . . .
Find the reduced tendon stress immediately after transfer due to elastic shortening.
Ag gross area of cross section 762 in2
Aps area of prestressing steel 6.732 in2
em average prestressing steel eccentricity at midspan 18.38 in
E w f kipsftci c ci= ′ =
33 000 33 000 0 151 53, , ..
11 5
6 0 4696.
. ksi ksi=
A Eq. 5.4.2.4-1
Eci modulus of elasticity of concrete at transfer 4696 ksi
Ep modulus of elasticity of prestressing tendon 28,500 ksi
fcgp concrete stress at center of gravity of prestressing
tendons due to prestressing force immediately after
transfer and self-weight of member at section of
maximum moment ksi
fpbt stress immediately prior to transfer ksi
fpj stress at jacking ksi
fpu specified tensile strength of prestressing steel 270 ksi
Ig moment of inertia of the gross concrete section 212,450 in4
Mg midspan moment due to member self-weight 632 ft-kips
MD midspan moment due to the slab dead load 600 ft-kips
Find the prestress loss due to elastic shortening, ΔfpES.
A Art. 5.9.5.2.3a-1
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∆fEE
fpESp
cicgp=
Find the initial prestress before transfer, but after the changes due to the elastic deformations of the section, fpbt, it is,
A Com. C5.9.5.2.3a
fpbt = 0.9 fpj
fpj = 0.75 fpu = (0.75)(270 ksi) = 203 ksi
A Tbl. 5.9.3-1
fpbt = (0.9)(203 ksi) = 182.3 ksi
To avoid iteration, alternately the loss due to elastic shortening may be determined
A Eq. C5.9.5.2.3a-1
∆fA f I e A e M A
A I e ApES
ps pbt g m g m g g
ps g m g
=+ −( )
+( )
2
2 ++
A I E
Eg g ci
p
=
( )( ) + ( )6 732 182 3 212 450 18 38 72 4 2. . , .in ksi in in 662
18 38 632 762
6 7
2
2
in
in f ips in
( )( )−( )( )( ).
.
t-k
332 212 450 18 38 762
7
2 4 2 2in in in in( ) ( ) + ( ) ( )( )+
, .
662 212 450 469628 500
2 4in in ksiksi
( )( )( )
,,
= 15.77 ksi
The reduced prestress force after transfer is,
P = (fpbt –ΔfpES)(Aps) = (182.3 ksi – 15.77 ksi)(6.732 in2)
= 1121.1 kips
179Design Examples
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Concrete stress at top fiber is
f PA
PeS
MSt
g
m
nc top
g
nc top= − + −
, ,
= − + ( )( )1121 1762
1121 1 18 387692
. . .kipsin
kips in22
632 12
76923 3in
f ips inft
in−
( )
t-k
= –1.47 ksi + 2.68 ksi – 0.98 ksi
= 0.23 ksi (tension) < fti = 0.563 ksi [OK]
Concrete stress in bottom fiber is
f PA
PeS
MSt
g
m
nc bottom
g
nc bottom= − − +
, ,
= − − ( )( )1121 1762
1121 1 18 389082
. . .kipsin
kips in77
632 12
90873 3in
f s inft
in+
( )
t-kip
= –1.47 ksi – 2.27 ksi + 0.83 ksi
= –2.91 ksi (compression) < fci = 3.6 ksi [OK]
Please see Figure 2.68.
+ =+
PAB
= –1,470
–
PAB
= –1,470
– PAB
– PemS
= –2,270–
PemS
PemS
PemS
–
= +2,680
+
MgS
MgS
MgS
MgS
= +830 = –2,910
+
PAg
– +
= +230
–
+
= –980
–
FIGURE 2.68Concrete stresses at midspan at release of prestress for girder I-beam.
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Step 5: Determine Final Concrete Stresses at Midspan after All Losses (Except Friction)
Find losses due to long-term shrinkage and creep of concrete, and low relax-ation strand.
A Art. 5.4.2.3.1; Tbl. 5.9.3-1
Prestressing steel stress immediately prior to transfer
fpbt = 0.75 fpu, not including the elastic deformation. Thus, fpi = fpj 203 ksi
H average annual ambient relative humidity 70% (for central NY)
A Fig. 5.4.2.3.3-1
ΔfpR an estimate of relaxation loss taken as 2.4 ksi for low relaxation strand, 10.0 ksi for stress relieved strand, and in accordance with manufacturer’s recommendation for other types of strand.
A Art. 5.9.5.3
Determine time-dependent losses:
A Art. 5.9.5.3
The correction factor for relative humidity of the ambient air is
A Eq. 5.9.5.3-2
γh = 1.7 – 0.01 H = 1.7 – (0.01)(70%) = 1.0
The correction factor for specified concrete strength at time of prestress transfer to the concrete member is,
A Eq. 5.9.5.3-3
γ stcif
=+ ′5
1
=+
=51 6 0
0 833.
.ksi
The long-term prestress loss due to creep of concrete, shrinkage of con-crete, and relaxation of steel is,
A Eq. 5.9.5.3-1
181Design Examples
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∆ ∆ff A
AfpLT
pi ps
gh st j st pR= + +10 0 12 0. .γ γ γ γ
=( )( )
( )10 0
203 6 732762
1 0 0 82
3..
. .ksi in
in333 12 0 1 0 0 833 2 4( ) + ( )( ) +. . . . ksi
= 25.4 ksi
The total loss for pretensioned members is
A Eq. 5.9.5.1-1ΔfpT = ΔfpES + ΔfpLT
= 15.77 ksi + 25.4 ksi = 41.17 ksi
The effective steel prestress after losses is
fse = fpi (= fpj) – ΔfpT = 203 ksi – 41.7 ksi = 161.3 ksi
The effective prestress force after all losses, Pe, is
Pe = fseAps = (161.3 ksi)(6.732 in2) = 1085.87 kips (1,085,870 lbf)
Determine final concrete stresses at midspan after the total losses.For the basic beam section at the beam’s bottom fiber,
− − ++P
AP e
SM MS
e
g
e m
nc bottom
g D
nc bottom, ,
= − − ( )(1 085 870762
1 085 870 18 382
, , , , .lbfin
lbf in))
+ +
9087
632 6007692
3
3
in
ips f psin
ft-k t-ki
12 1000inft
lbfkip
= –1994.4 psiwhere:Mg = moment due to beam weightMD = moment due to slab dead load
At the basic beam’s top fiber, the final concrete stresses at midspan are
− + −+P
AP eS
M MS
e
g
e m
nc top
g D
nc top, ,
= − + ( )(1 085 870762
1 085 870 18 382
, , , , .lbfin
lbf in))
− +
7692
632 6007692
3
3
in
s f psin
ft-kip t-ki
12 1000inft
lbfkip
= –752.3 psi
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Moment due to the superimposed dead loads, Ms, consists of the superim-posed dead load, ws, the parapet/curb load (0.506 kip/ft), distributed equally to the 5 girders, plus the load of the 3 in future wearing surface,
w
kipft
girdersin ft
ins =
( )
+ ( )0 506 2
53
12
.
( )0 14 7 53. .kips
ftft
= 0.4649 kips/ft
M w Lkips
ftft
ss= =
( )2
2
8
0 4649 80
8
.
= 371.9 ft-kips
For composite section at the beam base,
M M
Sf s f ss LL IM
bc
+ = ++ 371 9 1364 3814
. .,
t-kip t-kip5538 0
12 10003. ininft
lbfkip
= 1433 psi
At slab top,
− + = ++M MS
f ps f ss LL IM
tc
371 9 1364 3824
. .t-ki t-kip,, .476 6
12 10003ininft
lbfkip
= –851.2 psi
See Figure 2.69.
Check the allowable concrete stresses at Service I Limit State load com-bination after all losses.
A Art. 5.9.4.2
For compression at girder top,A Tbl. 5.9.4.2.1-1
fcs = 0.45 f′c = (0.45)(6.5 ksi) = 2.93 ksi < –1311.1 ksi [OK]
For tension at girder bottom,A Tbl. 5.9.4.2.2-1
183Design Examples
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f f ksi ksits c= ′ = =0 19 0 19 6 5 0 484. . . . > –0.5614 ksi [OK]
Design Example 7: Flexural and Transverse Reinforcement for 50 ft Reinforced Concrete Girder
Situation
BW barrier weight (curb, parapet, and sidewalk) 0.418 kips/ftf′c,beam beam concrete strength 4.5 ksif′c,deck deck concrete strength 3.555 ksiDW future wearing surface load 2 inL bridge span 50 ftLL live load HL-93Es modulus of elasticity of reinforcing steel 29,000 ksify specified minimum yield strength of reinforcing steel 60 ksiS girder spacing 8.0 ftts structural deck thickness 8.0 inwc concrete unit weight 150 lbf/ft3
Please see Figure 2.70.
With deadloads and prestressloads on basic beam
With superimposeddead loads plus
live loads oncomposite beam
Final Stress
–561.4+1433–1994.4
–752.3 –558.8 –1311.1
–851.2–851.2 = – 709.31.2
Slab
+
Note: Units are in lbf/in2; “–” Indicates compression.
558.8 = – 4651.2
=
FIGURE 2.69Final concrete stresses at midspan after losses.
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Requirements
Determine the flexural and transverse reinforcement for the reinforced con-crete girder described.
Solution
The effective flange width of a concrete deck slab may be taken as the tribu-tary width perpendicular to the axis of the member.
A Art. 4.6.2.6.1
be = 8 ft or 96 in
Step 1: Calculate the Beam Properties
The modulus of elasticity for the deck isA Eq. 5.4.2.4.1
E w f kipsftdeck c c= ′ = ( )
33 000 33 000 0 151 5
3, , ..
1 5
3 555.
. ksi
= 3615 ksi
The modulus of elasticity for the beam is
E w f kipsftdeck c c= ′ = ( )
33 000 33 000 0 151 5
3, , ..
1 5
4 5.
. ksi
= 4067 ksi
3 ft
40 in
43 ft46 ft
20 in10 in3 ft
26 in
1 ft 6 in
20 ft20 ft
5@ 8 ft = 40 ft
8 in deck
1 ft 6 in
FIGURE 2.70Reinforced concrete girder design example.
185Design Examples
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The modular ratio is
A Eq. 4.6.2.2.1-2
n EE
ksiksi
deck
beam= = =3615
40670 889.
The transformed effective deck width is
bet = (S)(n) = (8 ft)(0.889)(12 in/ft) = 85.34 in
The area of the T-beam is
A = (40 in)(20 in) + betts = 800 in + (85.34 in)(8 in) = 1482.72 in2
= 10.30 ft2
From bottom, the center of gravity of T-beam yb = ΣAy/ΣA, is
y in in in in in inb = + +( )( )( ) ( . )( )(40 20 20 85 34 8 40 4 iin
in in in in)
( )( ) ( . )( )40 20 85 34 8+
= 31.05 in from bottom
Moment of inertia of T-beam about the center of gravity is
I I bd Adg = = +3
3
12
I in in in ing =
( )( ) + ( )( )1
1220 40 20 40 31 053 . iin in
in inin in
−( )
+ ( )( ) + ( )( )
20
85 34 812
85 3 8
2
3.. 444 31 05 2in in−( ).
= 322,430 in4
= 15.55 ft4
Step 2: Calculate the Dead Load Effects
A Art. 3.3.2
The weight of the structural components is (see Figure 2.71)
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DC A w ft lbfftT beam c− = ( )( ) = ( )
=10 30 15023. 11545 lbf
ft
Permanent loads such as barriers may be evenly distributed across all beams. The weight of the barriers is
wBW
no of beams
lbfft
DC barrier, .= ( ) =
( )
2 2 418
=6
139 33. lbfft
wDC,total = wbeam + wbarrier = 1545 lbf/ft + 139.33 lbf/ft
= 1684.3 lbf/ft
The weight of the wearing surface and utilities is
wDW w L
no of beams
in ftin
DWc= ( ) =
( )
.
2 112
150 llbfft
ftlbfft
( )
=43
6179 2.
The unfactored maximum moments due to dead loads are
c.g. ofgirder
+
c.g. slab & girder
yb = 31.05 in20 in
20 in
eg = 24 in
8 in
bet = 85.34 in
40 in
FIGURE 2.71Girder section with area transformed deck slab.
187Design Examples
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M w Llbfft
ftDC
DC total= =
( )
,.2
2
8
1684 3 50
8
= 5.26 × 105 ft-lbf
M w Llbfft
ftDW
FWS= =
( )2
2
8
179 2 50
8
.
= 0.56 × 105 ft-lbf
The unfactored maximum moments occur at midspan.The unfactored maximum shears due to dead loads are
V w Llbfft
ftDC
DC total= =
( )
,.
2
1684 3 50
2
= 42,107 lbf
V w Llbfft
ftDW
DW= =
( )
2
179 2 50
2
.
= 4,480 lbf
The unfactored maximum shears occur at reactions for simply supported beams.
Step 3: Calculate the Live Load Effects
A Art. 3.6.1.2
The design live load, HL-93, shall consist of a combination of
• Design truck (HS-20) or design tandem and
• Design lane load
Dynamic load allowance, IM, is 33% for all limit states other than fatigue.
A Tbls. 3.6.2.1-1; 4.6.2.2.2b-1 or Appendix A
188 Simplified LRFD Bridge Design
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Calculate the live load distribution factor for moment, DFM, for interior beams of a typical cross section, e,
A Tbl. 4.6.2.2.1-1
The distance between the centers of gravity of the slab and the beam is
e in inin
ft
ftg = + =20 412 2
The stiffness parameter is
Kg = n(I + Aeg2) = (0.889)(15.55 ft4 + (10.30 ft2)(2 ft)2) = 50.44 ft4
= 1.046 x 106 in4 (10,000 ≤ Kg ≤ 7 x 106 in4) [OK]
A Eq. 4.6.2.2.1-1
Calculate the live load distribution factor for moment if the following are true.
A Tbl. 4.6.2.2.2b-1 or Appendix A
3.5 < S < 16 [OK]
4.5 < ts < 12 [OK]
20 < L <240 [OK]
Nb > 4 [OK]
The live load distribution factor for moment for interior beams with one design lane loaded is
DFM S SL
KLtsig
s= +
0 06
14 12
0 4 0 3
3.. .
0 1.
= +
×0 06 8
14850
1 05 100 4 0 3
... .
ft ftft
66 4
3
0 1
12 50 8
inin
ftft in
( )( )
.
= 0.581
189Design Examples
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The live load distribution factor for moment for interior beams with two or more design lanes loaded is
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
= +
×0 075 8
9 5850
1 050 6 0 2
..
.. .ft ft
ft
11012 50 8
6 4
3
0 1in
inft
ft in
( )( )
.
= 0.781
Calculate the live load distribution factor for moment for exterior beams with two or more design lanes loaded.
A Tbl. 4.6.2.2.2d-1 or Appendix B
The distance from the centerline of the exterior web of the exterior beam to the interior edge of the curb or traffic barrier, de, is
de = 18 in or 1.5 ft
The correction factor for distribution is
e dft
ftft
e= + = + =0 779 1
0 77 1 59 1
0 935..
. ..
.
Because –1.0 < de < 5.5, the live load distribution factor for moment for exte-rior beams with two or more design lanes loaded is
DFMme = (e)(DFMmi) = (0.935)(0.781) = 0.730
Calculate the live load distribution factor for moment for exterior beams with one design lane loaded using the lever rule. Please see Figure 2.72.
A Tbl. 4.6.2.2.2d-1; A Tbl. 3.6.1.1.2-1; C3.6.1.1.2
ΣM@hinge = 0
02
7 52
1 5 8= ( ) + ( ) − ( )P ft P ft R ft. .
R = 0.563 P
190 Simplified LRFD Bridge Design
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When the lever rule is used, the multiple presence factor, m, must be applied.
A Art. 3.6.1.1.2
The live load distribution factor for moment for exterior beams with one design lane loaded is
DFMse = (m)(0.563) = (1.2)(0.563) = 0.676
DFMmi = 0.781 for moment for interior beam [controls]
The live load distribution factor for shear for interior beams with two or more design lanes loaded is
A Tbl. 4.6.2.2.3a-1 or Appendix C
DFV Sft
Sft
ftft
fmi = + −
= + −0 212 35
0 2 812
82
. . ttft35
2
= 0.814
The live load distribution factor for shear for interior beams with one design lane loaded is
DFV Sft
ftftsi = + = +0 36
250 36 8
25. .
= 0.68
20 in10 in
2 ft 1.5 ft1.5 ft
P2
Hingeassumed
6 ft
3 ft
8 ft
de = 1.5 ft
P2
FIGURE 2.72Lever rule for distribution factor for exterior girder moment with one lane loaded.
191Design Examples
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Calculate the live load distribution factor for shear for exterior beams with two or more design lanes loaded.
A Tbl. 4.6.2.2.3b-1 or Appendix D
The distance from the centerline of the exterior web of the exterior beam to the interior edge of the curb or traffic barrier, de, is 18 in or 1.5 ft.
The correction factor for distribution is
e dft
ftft
e= + = + =0 610
0 6 1 510
0 75. . . .
Because –1.0 < de < 5.5, the live load distribution factor for shear for exterior beams with two or more design lanes loaded is
DFVme = (e)DFVint = (0.75)(0.814) = 0.610
The live load distribution factor for shear for exterior beams with one design lane loaded using the lever rule is the same as for the moment DFMse. Therefore,
DFVsi = 0.676
DFVmi = 0.814 for shear for interior beams controls
See Table 2.18.
Step 4: Calculate Truck Load (HS-20) and Lane Load Moments and Shears
A Art. 3.6.1.2
Calculate the unfactored flexural moment due to design truck load. (Refer to Design Example 1 and Figure 2.73.)
TABLE 2.18
Summary of Distribution Factors
Girder DFM DFV
Interior, two or more lanes loaded 0.781a 0.814a
Interior, one lane loaded 0.581 0.680Exterior, two or more lanes loaded 0.730 0.610Exterior, one lane loaded 0.676 0.676
a Controlling distribution factors.
192 Simplified LRFD Bridge Design
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ΣM@A = 0
Rlbf ft lbf ft
B = ( )( ) + ( )( ) +8000 11 32 000 25 32 000, , llbf ftft
( )( )3950
= 42,720 lbf
The unfactored flexural moment at midspan due to design truck load is
Mtr = (–32,000 lbf)(14 ft) + (42,720 lbf)(25 ft)
= 6.2 x 105 lbf-ft per lane
(Also see Design Example 1.)The unfactored flexural moment at midspan due to design tandem load
is
Mtandem = 575,000 lbf-ft per lane
(See Design Example 1)Calculate the unfactored flexural moment due to lane load.
The ultimate moment due to live load of 640 lbf/ft occurs at the center of simply supported beam. See Figure 2.74.
A Art. 3.6.1.2.4
M w Llbfft
ftLL
ln = =
( )2
2
8
640 50
8
= 2.0 × 105 lbf-ft per lane
25 ft25 ftRA RBCL
32 kips14 ft14 ft11 ft 11 ft32 kips8 kips
A B
C
FIGURE 2.73Design truck (HS-20) load position for the maximum moment at midspan.
193Design Examples
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Design lane load is not subjected to a dynamic load allowance.A Art. 3.6.1.2.4
The total unfactored flexural moment due to live loads per lane is (See Figure 2.75)
MLL+IM = Mtr(1 + IM) + Mln = (6.2 × 105 lbf-ft)(1.33) + 2.0 × 105 lbf-ft
= 10.246 × 105 lbf-ft per lane
The total unfactored live load moment per girder is
Mtotal = MLL+IM(DFM)
= (10.2 × 105 lbf-ft)(0.781)
= 7.966 × 105 lbf-ft per girder
Calculate the unfactored shear due to design truck load.
ΣM@B = 0
RA RBL = 50 ft
wLL = 640 lbf/ft
A BC
FIGURE 2.74Design lane load moment at midspan.
RA RB50 ft
22 ft14 ft8 kips32 kips32 kips
14 ft
A B
FIGURE 2.75Design truck load (HS-20) load position for the maximum shear at support.
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Rlbf ft lbf ft
A = ( )( ) + ( )( ) +8000 22 32 000 36 32 000, , llbf ftft
( )( )5050
= 58,560 lbf
V = RA = 58,560 lbf
The unfactored shear due to design tandem load is
Vtandem = 48,000 lbf (See Design Example 1.)
The unfactored shear due to design truck load is
Vtr = V(1 + IM) = (5.856 × 104 lbf)(1.33)
= 7.79 × 104 lbf
Calculate the unfactored shear due to lane load.
The ultimate shear due to uniform live load of 640 lbf/ft occurring at the end of a simply supported beam is,
V R w Llbfft
ftlane A
LL= = =
( )
2
640 50
2
= 1.6 × 104 lbf per lane
Dynamic effects are not taken into consideration for design lane loading. Please see Figure 2.76.
A Art. 3.6.1.2.4
RA RBL = 50 ft
wLL = 640 lbf/ft
A B
FIGURE 2.76Design lane load position for the maximum shear at support.
195Design Examples
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The total unfactored shear per lane is
VLL+IM = Vtr + Vln = (7.79 × 104 lbf) + (1.6 × 104 lbf)
= 9.39 × 104 lbf per lane
The total unfactored shear per interior girder is
Vtotal = VLL+IM(DFV) = (9.39 × 104 lbf)(0.814) = 7.64 × 104 lbf
Step 5: Calculate the Strength I Limit State
The factored moment for Strength I Limit State is
Mu = 1.25 MDC + 1.5 MDW + 1.75 Mtotal
= (1.25)(5.26 × 105 ft-lbf) + (1.5)(0.56 × 105 ft-lbf) + (1.75)(7.996 × 105 ft-lbf)
= 21.4 × 105 ft-lbf
The factored shear for Strength I Limit State is
Vu = 1.25 VDC + 1.5 VDW + 1.75 Vtotal
= (1.25)(4.21 × 104 lbf) + (1.5)(0.448 × 104 lbf) + (1.75)(7.64 × 104 lbf)
= 19.3 × 104 lbf = 193.0 kips
Step 6: Design the Reinforcement
Design the flexural reinforcement. Neglect compression reinforcement in the flange. See Figure 2.77.
A Art. 5.14.1.5.1c; C5.14.1.5.1c
The web thickness with 5 #11 bars in each layer and #4 stirrup bars, bw, is
bw = (2)(1 in) + (2)(0.5 in) + 0.5 db + (4)(1.5 db)
= (2)(1 in) + (2)(0.5 in) + 0.5(1.4 in) + (4)(1.5)(1.4 in)
= 18.4 in
To allow for extra room, use bw = 20 in.
196 Simplified LRFD Bridge Design
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Try using 10 no. 11 bars in two rows. The area of longitudinal reinforce-ment, As, is 15.6 in2.
A Art. 5.10.3.1; 5.12.3
Assuming no. 4 stirrups and a 1.0 in concrete cover (assuming epoxy-coated bars), the distance from the girder base to the centroid of the rein-forcement is
1 0 0 52
1 5 1 41 1 412
3 6. . . . . .in in d d in in inb
b+ + + = + + = 22 in (4 in)
The stress block factor, β1, isA Art. 5.7.2.2; 5.7.3.1.1
β1 0 85 0 05 4= − ′ −( ). . f ksic
= 0.85 – 0.05(4.5 ksi – 4 ksi)
= 0.825
The distance from the extreme compression fiber to the neutral axis is
cA ff b
in ksis y
c et=
′=
( )( )( )0 85
15 6 600 85 41
2
..
. .β 55 0 825 85 343 47
ksi inin( )( )( ) =
. ..
The depth of the equivalent rectangular stress block is
20 in 1.5 in10 #11 bars
No. 4 barstirrups
bet = 85.34 in
40 in
8 in
4 in
ds = 44 in
FIGURE 2.77Reinforcement details.
197Design Examples
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a = cβ1 = (3.47 in)(0.825) = 2.86 in
The nominal resisting moment isA Eq. 5.7.3.2.2-1
M A f d a in ksi inn s y= −
= ( )( ) −2
15 6 60 44 2 862. . iin ftin2
112
= 3320.5 ft-kips
The resistance factor, Φ, is 0.90.A Art. 5.5.4.2.1
The factored resisting moment, Mr, isA Eq. 5.7.3.2.1-1
Mr = ΦMn = (0.9)(3320.5 ft-kips)
= 2988.5 ft-kips > Mu = 2140.0 ft-kips [OK]
Step 7: Check the Reinforcement Requirements
Review the flexural reinforcement.
The maximum flexural reinforcement provision was deleted in 2005.A Art. 5.7.3.3.1
The minimum reinforcement requirement is satisfied if Mr (= ΦMn) is at least equal to the lesser of
A Art. 5.7.3.3.2
• Mr = ΦMn ≥ 1.2 Mcr
or
• Mr ≥ (1.33)(the factored moment required by the applicable strength load combination)
where:
Sc section modulus for the tensile extreme of the gross section in3
fr modulus of rupture of concrete, 0.37√f′c ksiA Art. 5.4.2.6
f′c girder concrete strength 4.5 ksi
198 Simplified LRFD Bridge Design
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yb center of gravity of T-beam from tensile extreme face 31.05 in
Ig moment of inertia of the gross concrete section 322,430 in4
M S fIy
f inincr c r
g
br= =
=
322 43031 05
4,. ( )0 37 4 5. . ksi
= 8150.45 kip-in = 679.2 kip-ft
A Eq. 5.7.3.3.2-1
1.2 Mcr = (1.2)(679.2 kip-ft) = 815.04 kip-ft, or
1 33 1 33 21 40 10 11000
5. . .M lb t kiplbfu = ( ) ×( )
f-f
= 2846.20 kip-ft
1.2 Mcr = 815.04 kip-ft [controls]
Mr = ΦMn = 2988.5 kip-ft > 1.2 Mcr = 815.04 kip-ft [OK]
The minimum reinforcement required is satisfied with 10 no. 11 bars as tension reinforcement.
Review the shear reinforcement.
A Art. 5.8.2.9
For simplicity, shear forces at end supports will be considered.
Find the factored shear for Strength I Limit State, Vs.
The effective shear depth need not be taken to be less than the greater of 0.9 de or 0.72 h.
d d a in in inv s= − = − =2
44 2 862
42 57. . [controls]
The minimum dv is
dv = 0.9 de = (0.9)(44 in) = 39.6 inor dv = 0.72 h = (0.72)(48 in) = 34.5 in
See Figure 2.78.
199Design Examples
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The transverse reinforcement shall be provided whereA Art. 5.8.2.4
Vu > 0.5 ΦVc
A Eq. 5.8.2.4-1
β is 2.0 when Q = 45° for diagonal cracks.A Art. 5.8.3.4.1; A Eq. 5.8.3.3-3
V f b d ksi inc c v v= ′ = ( )( ) ( )0 0316 0 0316 2 0 4 5 20. . . .β 442 57. in( )
= 114.1 kips
0.5 ΦVc = (0.5)(0.90)(114.1 kips)
= 51.35 kipsA Eq. 5.8.2.4-1
Vu = 193 kips > 0.5 ΦVc (= 51.35 kips)
Therefore, the transverse reinforcement is needed at end support points.A Art. 5.8.2.4
The nominal shear resistance, Vn, is the lesser ofA Art. 5.8.3.3
Vn = Vc + Vs
Vn = 0.25 f′c bvdv
bv = 20 in
a = 3.23 in
C
40 indv = 42.57 in
T
h = 48 in
8 in
ds = de = 44 in
4 in
FIGURE 2.78Review of shear reinforcement.
200 Simplified LRFD Bridge Design
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Vs is the shear resistance provided by shear reinforcement.
Vn = 0.25 f′c bvdv = (0.25)(4.5 ksi)(20 in)(42.57 in) = 957.8 kips
Let the factored shear resistance, Vr, be equal to the factored shear load, Vu
A Eq. 5.8.2.1-2
Vr = ΦVn = Vu
V V kips kipsnu= = =
Φ193 2
0 90214 4.
.. [controls]
The shear required by shear reinforcement, Vs, is found as follows.
A Eq. 5.8.3.3.1
Vn = Vc + Vs
V V V V V kips kipss n cu
c= − = − = −Φ
214 40 90
114 1..
.
= 124.1 kips
The angle of inclination of stirrups to longitudinal axis, α, is 90° and the angle of inclination of diagonal stress, θ, is 45°. Therefore, the shear resis-tance by shear reinforcement, Vs, and the spacing of stirrups, s, are calculated as follows.
Eq. C5.8.3.3-1, A Eq. 5.8.3.3.4; A Art. 5.8.3.4.1
VA f d
ssv y v=
+( )cot cot sinθ α α
Using two #4 bar stirrups,
sA f d
Vin ksi inv y v
s= =
( )( )( )( )cot . .θ 2 0 20 60 42 572 11 0124 1
..
( )kips
= 8.2 in [Try s = 8.0 in.]
The minimum transverse reinforcement required is
A Eq. 5.8.2.5-1
201Design Examples
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A f b sf
ksiin in
v cv
y= ′ = ( )( )0 0316 0 0316 4 5
20 860
. . .kksi
in
= 0 18 2.
The provided area of the shear reinforcement within the given distance, Av, of two times 0.20 in2 with no. 4 stirrups at 8 in spacing is OK.
Summary
For flexural reinforcement, use 10 no. 11 bars in 2 rows. For shear reinforce-ment, use no. 4 stirrups at 8 in spacing.
Design Example 8: Determination of Load Effects Due to Wind Loads, Braking Force, Temperature Changes, and Earthquake Loads Acting on an Abutment
A two-lane bridge is supported by seven W30 × 108 steel beams with 14 in × 0.6 in cover plates. The overall width of the bridge is 29.5 ft and the clear roadway width is 26.0 ft with an 8 in concrete slab. The superstructure span is 60.2 ft center to center on bearings (the overall beam length is 61.7 ft). It is 3.28 ft high including the concrete slab, steel beams, and 0.12 ft thick bearing.
The abutment has a total height of 16 ft and a length of 29.5 ft placed on a spread footing. The footing is placed on a gravel and sand soil.
Solution
Determine other load effects on the abutment.
1. Wind Loads
It is noted that the transverse wind loads (lateral to girder) in the plane of abut-ment length will be neglected because the abutment length is long (i.e., 29.5 ft).
Therefore, wind loads in longitudinal directions (perpendicular to the abutment wall face) will be considered.
1.A Wind Pressures on Structures, WS
A Art. 3.8.1.2
The skew angle as measured from a perpendicular to the beam longitudinal axis is 30°. See Figures 2.79 through 2.81.
202 Simplified LRFD Bridge Design
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Total depth away from abutment support = hparapet + tdeck + dgirder + tbearing,
=
+
+32 112
8 112
2 49in ftin
in ftin
ft B. eeam With No Cover Plate
ft Cover Plate
( )
+ ( )0 05.
= 5.87 ft
Total depth at abutment support = hparapet + tdeck + dgirder + tbearing,
=
+
+32 112
8 112
2 49in ftin
in ftin
ft B. eeam With No Cover Plate
ft Bearing Plate
( )
+ ( )0 12.
= 5.94 ft
61.7 ft3.28 ft to
the roadway60.2 ft
12 ft
2 ft
16 ftAbutments are 16 fthigh and 29.5 ft long
4 ft 11 ft 11 ft
FIGURE 2.79Abutment structure 16 ft in height and 29.5 ft in width.
29 ft 6 in
Top ofabutment
29 ft 6 in
26 ft 0 in (roadway)
6 at 4 ft 3 in
1 ft 9 in
CL2 ft
FIGURE 2.80Two-lane bridge supported by seven W30 × 108 steel beams at 29.5 ft wide abutment.
203Design Examples
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Overall beam length = 61.70 ft
Bearing to bearing length = 60.2 ft
A base design wind velocity at 30 ft height is,A Art. 3.8.1.1
V30 = VB = 100 mph
Inasmuch as the abutment is less than 30 ft above ground level, the design wind velocity, VDZ, does not have to be adjusted.
VDZ = VB
Design wind pressure, PD, isA Art. 3.8.1.2.1
PD =
P VVB
DZ
B
2
2.49 ftbeam
(no coverplate)
2.49 ft
0.05 ftcover
2.54 ft
5.87 ft 3.28 ft
0.12 ft bearing
W30 × 108
Slab8 in
32 in
Away fromabutment
12 in 2 in
7 in
1 ft 7 in
Atabutment
Curband
parapet
10 in
3 in
4 ft
FIGURE 2.81Steel beams at abutment and away from abutment.
204 Simplified LRFD Bridge Design
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where:PB = base wind pressure specified in AASHTO [Tbl. 3.8.1.2.1-1] (kips/ft2)
Base wind pressure, PB, with skew angle of 30° is 0.012 kips/ft2 for longi-tudinal load.
A Tbl. 3.8.1.2.2-1
The design longitudinal wind pressure, PD, is
PD =
0 012 100
1002
2
. kipsft
mphmph
= 0.012 kips/ft2
The total longitudinal wind loading, WStotal, is
WStotal = (61.70 ft)(5.87 ft)(0.012 ksf)
WStotal = 4.35 kips
The longitudinal wind force from super structure, WSh, must be transmit-ted to the substructure through the bearings as follows:
4.35 kips wind loads from superstructure acts at its mid-depth (at the abut-ment) which is (5.87 ft – 0.05 ft cover plate)/2 + 0.12 ft bearing = 3.03 ft at the top of the abutment.
The longitudinal (horizontal) wind loading at the top of the abutment is
WSh = 4.35 kips/29.5 ft abutment length
WSh = 0.15 kips/ft abutment width
The vertical wind loading at the top of the abutment, WSv, is the reaction (see Figure 2.82)
(4.35 kips)(3.03 ft)/60.2 ft = 0.22 kips up or down.
Along the abutment width 29.5 ft, the vertical wind loading is,
WSv = 0.22 kips/29.5 ft
= 0.007 kips/ft abutment width (negligible)
205Design Examples
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1.B Aeroelastic Instability
A Art. 3.8.3
Span
Abutment Widthftft
= =60 229 5
2 04..
.
Span
Girder Depthftft
= =60 25 87
10 2..
.
The ratios are less than 30.0. Therefore, the bridge is deemed to be not wind sensitive.
1.C Wind Pressures on Vehicle Live Load, WL
A Art. 3.8.1.3, A Tbl. 3.8.1.3-1
WL = (0.024 kips/ft for the skew angle of 30°)(61.7 ft)
= 1.48 kips acting at 6.0 ft above the deck
Or 1.48 kips will be acting at (6 ft + 3.28 ft) = 9.28 ft above the abutment top.This 1.48 kips longitudinal force is transmitted to the abutment through
the bearings as follows and as shown in Figure 2.83:The reaction at the fixed abutment is
1 48 9 28
60 20 228
. ..
.kips ft
ftkips( )( ) =
The longitudinal (horizontal) wind loading at the top of abutment due to vehicle live load is
WLh = =1 4829 5
0 05..
.kipsft
kipsft abutment width
60.20 ft
4.35 kips 0.12 ftbearing
0.22 kips
3.03 ft
Girder
0.22 kips
FIGURE 2.82Wind loads on abutment transmitted from superstructure.
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The vertical wind loading at the top of the abutment due to vehicle live load is
WLv = =0 22829 5
0 0077..
.kipsft
kipsft abutment width
(negligible)
1.D Vertical Wind Pressure
A Art. 3.8.2; Tbl. 3.4.1-1
A vertical upward wind force shall be applied only for Strength III and Service IV Limit States. Therefore, the vertical upward wind force is not applicable to Strength I Limit State.
1.E Wind Forces Applied Directly to the Substructure, Wsub
A Art. 3.8.1.2.3
The design wind pressure, PD, is
PD =
P VVB
DZ
B
2
where:PB = base wind pressurePB = 0.04 kips/ft2
A Art. 3.8.1.2.3
1.48 kips
0.12 ft9.28 ft
0.2280 kips
60.2 ft0.228 kips
Bearing
Girder
FIGURE 2.83Wind loads on abutments transmitted from vehicle live load.
207Design Examples
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PD =
P VVB
DZ
B
2
PD =
0 04 100
1002
2
. kipsft
mphmph
PD = 0.04 kips/ft2
The wind loading of 0.04 kips/ft2 will act, conservatively, perpendicular to the exposed 12 ft stem of the abutment, and it will act at 10 ft (= 12 ft/2 + 4 ft) above the base of the footing.
WSsub = (0. 04 ksf)(12 ft) = 0.48 kips/ft abutment width
2. Braking Force, BR
A Art. 3.6.4
The braking force shall be taken as the greater of :
• 25% of the axle weights of the design truck or design tandem or• 5% of the design truck plus lane load or• 5% of the design tandem plus lane load
The static effects of the design truck or tandem for braking forces are not increased for dynamic load allowance.
A Art. 3.6.2.1
The total braking force is calculated based on the number of design lanes in the same direction.
A Art. 3.6.1.1; 3.6.4
In this example, it is assumed that all design lanes are likely to become one-directional in the future.
The braking force for a single traffic lane is as follows:
BRtruck = 25% of design truck = (0.25)(32 kips + 32 kips + 8 kips)
= 18.0 kips [controls]
BRtandem = 25% of the design tandem = (0.25)(25 kips + 25 kips)
= 12.50 kips
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BRtruck+lane = 5% of the design truck plus lane load
= 0.05[(32 kips + 32 kips + 8 kips) + (0.64 kips/ft x 60.2 ft)]
= 7.45 kips
BRtandem+lane = 5% of the design tandem plus lane load
= 0.05[(25 kips + 25 kips) + (0.64 kips/ft x 60.2 ft)]
= 4.43 kips
The braking forces act horizontally at a distance 6 ft above the roadway surface in either longitudinal direction. The multiple presence factor in AASHTO [3.6.1.1.2] shall apply.
A Art. 3.6.4
For the number of loaded lanes of 2, multiple presence factor, m, is 1.0.A Tbl. 3.6.1.1.2-1
Maximum braking force,
BRmax = BRtruck = 18.0 kips
The longitudinal (horizontal) braking force is transmitted to the abutment as follows.
The longitudinal (horizontal) force at the top of the abutment due to brak-ing force is (see also Figure 2.84):
BRhor = =18 029 5
0 61..
.kipsft
kipsft abutment width
2.77 kips
9.28 ft
2.77 kips
18.0 kips
FIGURE 2.84Forces on abutment from braking.
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The vertical reaction at the abutment is:
18 0 9 28 6 3 28. . .kips ft ft above the roadway ft× = +( )
660 22 77
..
ftkips=
The vertical force at the top of the abutment due to braking force is:
BRvert = =2 7729 5
0 094..
.kipsft
kipsft abutment width
3. Force Effect Due to Uniform Temperature Change
A Art. 3.12.2
Assume a moderate climate. The temperature range is 0°F to 120°F. Also, assume steel girder setting temperature Tset is 68°F.
A Tbl. 3.12. 2. 1-1
The thermal coefficient of steel is 6.5 × 10-6 in/in/°F.A Art. 6.4.1
Expansion thermal movement
Δexp = ×
°
° − °( )−6 5 10 120 68 60 2 126. .ininF
F F ft iinft1
= 0.244 in
Contraction thermal movement
Δcontr = ×
°
° − °( )−6 5 10 68 0 60 2 126. .ininF
F F ft in11 ft
= 0.32 in
The elastomeric bearing pad properties assumed are:
Shear modulus of the elastomer, G, isA Art. 14.7.5.2
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G = 0.095 ksi
0.08 ksi < 0.095 ksi < 0.175 ksi [OK]
Pad area, A, isA Art. 14.6.3.1; 14.7.2.3.1
A = 210.0 in2
Pad thickness = 3.5 in
Lateral (horizontal) loads due to temperature,A Eq. 14.6.3.1-2
H GAhbu
u
rt=
∆
where:Δu = shear deformationhrt = elastomer thickness
The load due to expansion is
Hrise = ( )( )
0 095 210 0 0 2443 5
2. . ..
ksi in inin
= 1.39 kips per bearing
Multiply Hrise by seven beam bearings and divide by the abutment length to determine the lateral load at the top of the abutment.
Htemp,rise =×( ) =
1 39 729 5
0 33.
..
kips bearingsft
kipsft abutmeent width
The load due to contraction is
Htemp,fall = ( )( )
0 095 210 0 0 323 5
2. . ..
ksi in inin
= 1.82 kips per bearing
The lateral (horizontal) load at the top of the abutment due to temperature fall is calculated in the same manner as for temperature rise.
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Htemp fall =×( ) =
1 82 729 5
0 43.
..
kips bearingsft
kipsft abutmeent width
controls
4. Earthquake Loads
A Art. 3.10; 3.10.9; 4.7.4.1; 4.7.4.2
First Step: Describe the type of bridge, number of spans, height of piers, type of foundations, subsurface soil conditions, and so on.
Second Step: Determine the horizontal acceleration coefficient, A, that is appropriate for bridge site (Art. 3.10.2; Fig. 3.10.2.1-1 and 3.10.2.1-2; Tbl. 3.10.3.1-1) and seismic zones (Art. 3.10.9; Art. 3.10.6). The values of the coefficients in the contour maps are expressed in percent. Thus, numerical values for the coef-ficient are obtained by dividing the contour values by 100. Local maxima and minima are given inside the highest and lowest contour for a particular region.
A Art. 3.10.2
Third Step: Determine the seismic performance zone for the bridge.A Art. 3.10.6
Fourth Step: Determine the importance category of the bridge.A Art. 3.10.5
Fifth Step: Determine the bridge site coefficient, which is based on soil pro-file types defined in Art. 3.10.3.1 through 3.10.7.1.
A Art. 3.10.3
Sixth Step: Determine the response modification factors (R factors), which reduce the seismic force based on elastic analysis of the bridge. The force effects from an elastic analysis are to be divided by the R factors.
A Art. 3.10.7; Tbl. 3.10.7.1-1; 3.10.7.1-2
Based on the information from the six steps, the level of seismic analysis required can be determined.
Single-Span Bridges
A Art. 4.7.4.2
Seismic analysis is not required for single-span bridges, regardless of seis-mic zone.
Connections between the bridge superstructure and the abutment shall be designed for the minimum force requirements as specified in AASHTO
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[3.10.9]. Minimum seat width requirements shall be satisfied at each abut-ment as specified in AASHTO [4.7.4.4].
Seismic Zone 1
A Art. 3.10.9.2, C3.10.9.2, 3.10.9
For New York City, the peak ground acceleration coefficient on rock (site class B), PGA, is 0.09.
A Fig. 3.10.2.1-1
Site Class B (Rock)A Tbl. 3.10.3.1-1
Value of site factor, fpga, for the site class B and PGA less than 0.10 is 1.0.A Tbl. 3.10.3.2-1
The peak seismic ground acceleration, As, isA Eq. 3.10.4.2-2
As = (fpga)(PGA) = (1.0)(0.09)
= 0.09 ≥ 0.05
The horizontal design connection force in the restrained directions shall not be less than 0.25 times the vertical reaction due to the tributary perma-nent load and the tributary live loads assumed to exist during an earthquake.
A Art. 3.10.9.2
The magnitude of live load assumed to exist at the time of the earthquake should be consistent with the value of γEQ for the extreme event I Limit State used in conjunction with Table 3.4.1-1. γEQ issue is not resolved and therefore used as 0.0.
A Comm. 3.10.9.2
Because all abutment bearings are restrained in the transverse direction, the tributary permanent load can be taken as the reaction at the bearing. Therefore, no tributary live loads will be considered.
A Art. 3.10.9.2
The girder vertical reaction, DL = 78.4 kips [previously known].
The maximum transverse horizontal earthquake load = 0.25 x 78.4 kips = 19.6 kips.
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The transverse (horizontal) earthquake loading at the top of the abutment is
EQh = =19 629 5
0 66..
.kipsft abutment
kipsft abutment wiidth
Please see Table 2.19 and Figure 2.85 for the summary of the forces. For the load combinations and load factors, refer to AASHTO Table 3.4.1-1 for the appropriate limit states.
TABLE 2.19
Summary of Forces
Component Description
Forces per ft of Abutment Width
(kips)
WSh Longitudinal (horizontal) wind loading 0.15WSv Vertical wind loading 0.007WLh Longitudinal (horizontal) wind loading due to vehicle
live load0.05
WLv Vertical wind loading due to vehicle live load 0.0077WSsub Wind loading perpendicular to exposed abutment stem 0.48BRhor Longitudinal (horizontal) force due to braking force 0.61BRvert Vertical force due to braking force 0.094Htemp, rise Lateral load due to temperature rise 0.33Htemp, fall Lateral load due to temperature fall 0.43EQh Transverse (lateral) earthquake loading 0.66
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WSh = 0.15 kips/ft
WSsub = 0.48 kips/ft
WLh = 0.05 kips/ft
WSv = 0.007 kips/ftWLv = 0.0077 kips/ftBRvert = 0.094 kips/ft
BRhor = 0.61 kips/ftHtemp,fall = 0.43 kips/ftEQh = 0.66 kips/ft
6 ft
12 ft
1 ft 2 in
2 ft 3 ft 2 ft 8 in 5 ft 4 in
2 ft
3.28 ft
FIGURE 2.85Summary of forces on abutment due to wind loads, braking forces, temperature changes, and earthquake loads.
215© 2008 Taylor & Francis Group, LLC
3Practice Problems
Practice Problem 1: Noncomposite 60 ft Steel Beam Bridge for Limit States Strength I, Fatigue II, and Service
Situation
L bridge span 60 ftLL live load HL-93FWS future wearing surface (bituminous concrete) 3 inf′c concrete strength 5.0 ksiADTT average daily truck traffic 250 fatigue design life 75 years New Jersey barrier 0.45 kips/ft eachw roadway width 43 ftwc dead load of concrete 150 lbf/ft3
ww dead load of wearing surface 140 lbf/ft3
(bituminous concrete)
Please see Figure 3.1.
Requirements
Review:
Flexural and shear resistance for Strength I Limit StateFatigue II Limit StateService Limit State Elastic deflection Flexure requirement to prevent permanent deflections
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Solution
Step 1: Review Section, W40 × 249
AISC (13th ed.) Tbl. 1-1
Please see Figure 3.2.
Deck slab, 8 in (7.5 in structural)Effective Flange Width of a concrete deck slab be taken as the tributary
width perpendicular to the axis of the member.A Art. 4.6.2.6.1
1.42 in
AISC 13th Ed. Table 1-1
0.75 in39.4 in
15.8 in
A = 73.3 in2
Fy = 50 kips/in2
Ix = 19,600 in4
Iy = 926 in4
Sx = 993 in3
Sy = 118 in3
FIGURE 3.2W40 × 249 properties.
3 ft
W40 × 249Fy = 50
3 ft
2 inhaunch
10 ft 10 ft
3 in FWS8 in slab
(7.5 in structural)
10 ft 10 ft
1.5 ft1.5 ft43 ft
roadway
kipsin2
FIGURE 3.1Cross section of noncomposite steel beam bridge.
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ts slab thickness 8 inbi effective flange width for interior beams inbe effective flange width for exterior beams inbw web width 0.75 inbf flange width of steel section 15.8 inS average spacing of adjacent beams 10 ft
The effective flange width for interior beams is
bi = S = 10 ft = 120 in
The effective flange width for exterior beams is
b ft inft
ft infte = ( )
+
3 12 102
12 = 96 in
Step 2: Evaluate Dead Loads
The structural components dead load, DC1, per interior girder is calculated as follows and as shown in Figure 3.3.
DC ininft
ininft
slab =
120
12
8
1200 150 1 03. . /kips
ftkips ft
=
DC ininft
ininft
haunch =
2
12
15 8
12
.
=0 150 0 0333. . /kipsft
kips ft
15.8 in
39.4 in
2 in
W40 × 249
8 in slab
10 ft = 120 in
FIGURE 3.3Dead loads for interior girder.
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Assuming 5% of steel weight for diaphragms, stiffeners, and so on,
DC kipsft
kipsftslab = + ( )
=0 249 0 05 0 249. . . 00 261. /kips ft
DC lbfft
roadway ws lace formstay-in-p =
7 2iidth
no of girderskipsft.
.
=
0 007 42
335
ft
= 0.06 kips/ft
The structural components dead load is
DC1 = DCslab + DChaunch + DCsteel + DCSIP forms
= 1.0 kips/ft + 0.033 kip/ft + 0.261 kips/ft + 0.06 kips/ft = 1.354 kips/ft
The shear for DC1 is
V wLkips
ftft
kipsDC1 2
1 354 60
240 62= =
( )
=.
.
The moment for DC1 is
M wLkips
ftft
DC1
22
8
1 354 60
8609 3= =
( )
=.
. ft--kips
The nonstructural dead load, DC2, per interior girder is calculated as
DC
kipsft
barriers
girdebarrier =
( )0 45 2
5
.
rrskips ft= 0 18. /
The shear for DC2 is
V wLkips
ftft
kipsDC2 2
0 18 60
25 4= =
( )
=.
.
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The moment for DC2 is
M wLkips
ftft
fDC2
22
8
0 18 60
881 0= =
( )
=.
. t-kiips
Total structural and nonstructural dead load, DCtot, for interior girders is
VDC,tot = VDC1 + VDC2 = 40.62 kips + 5.4 kips = 46.02 (46 kips)
MDC,tot = MDC1 + MDC2 = 609.3 ft-kips + 81.0 ft-kips = 690.3 ft-kips
The wearing surface dead load, DW, for interior girders is
DWFWS w L
no of beams
in ftin
FWSw= ( ) =
( )
.
3 112
1140 43
5
300 0 3
3lbfft
ft
beams
lbf ft k
( )
= =/ . iips ft/
The shear for DW is
V wLkips
ftft
kipsDW = =
( )
=2
0 3 60
29 0
..
The moment for DW is
M wLkips
ftft
f ipsDW = =
( )
=2
2
8
0 3 60
8135
.t-k
Find the structural components dead load, DC1, per exterior girder (see Figure 3.4).
DC ft ininft
kipsftslab = ( )
8 8
120 150 3.
= 0 80. /kips ft
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DC ininft
in ftinhaunch =
( )
2
1215 8 1
12.
=0 150 0 0333. . /kipsft
kips ft
DC kipsft
kipsftsteel =
+ ( )0 249 0 05 0 249. . .
= 0 261. /kips ft
DC lbf
ftroadway w
sta rmsy-in-place fo =
7 2iidth
no of girderskipsft.
.
0 007 432 fft
kips ft
( )
=
15
0 06. /
The structural components dead load, DC1, per exterior girder is
DC1 = DCslab + DChaunch + DCsteel + DCSIP forms
= 0.80 kips/ft + 0.033 kip/ft + 0.261 kips/ft + 0.06 kips/ft = 1.154 kips/ft
The shear for DC1, per exterior girder is
V wLkips
ftft
kipsDC1 2
1 154 60
234 62= =
( )
=.
.
15.8 in
3 ft 5 ft
2 in8 in slab
3 ft + 5 ft = 8 ft
5 ft
FIGURE 3.4Dead loads for exterior girder.
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The moment for DC1, per exterior girder is
M wLkips
ftft
fDC1
22
8
1 154 60
8519 3= =
( )
=.
. t--kips
The nonstructural dead load, DC2, per exterior girder is
DC
kipsft
barriers
girdebarrier =
( )0 45 2
5
.
rrskips ft= 0 18. /
The shear for DC2, per exterior girder is
V wLkips
ftft
kipsDC2 2
0 18 60
25 4= =
( )
=.
.
The moment for DC2, per exterior girder is
M wLkips
ftft
f pDC2
22
8
0 18 60
881= =
( )
=.
t-ki ss
The shear for the structural and nonstructural dead load, DCtot, for exterior girder is
VDC,tot = VDC1 + VDC2 = 34.62 kips + 5.4 kips = 40.0 kips
The moment for the structural and nonstructural dead load for exterior girder is
MDC,tot = MDC1 + MDC2 = 519.3 ft-kips + 81 ft-kips = 600.3 ft-kips
The shear for the wearing surface dead load, DW, for exterior girders is
V wLkips
ftft
kipsDW = =
( )
=2
0 3 60
29 0
..
The moment for the wearing surface dead load, DW, for exterior girders is
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M wLkips
ftft
f ipsDW = =
( )
=2
2
8
0 3 60
8135
.t-k
Please see Table 3.1.
Step 3: Evaluate Live Loads
The number of lanes (with fractional parts discounted) is
N w ftft
lane
lanesL = = =12
43
123 58. (3 lanes)
Find the longitudinal stiffness parameter, Kg.A Art. 4.6.2.2.1
The modulus of elasticity for the concrete deck isA Eq. 5.4.2.4-1
E E w f kipsftdeck c c c= = ′ = ( )33 000 33 000 0 151 5
3, , ..
=5 4286 8 2ksi kips in. /
The modulus of elasticity for the beam, Ebeam, is 29,000 kips/in2.The modular ratio between steel and concrete is
n EE
ksiksi
beam
deck= =29 000
4286 86 76,
..
The distance between the centers of gravity of the deck and beam is (see Figure 3.5)
eg = 19.7 in + 2 in + 4 in = 25.7 in
The moment of inertia for the beam, I (steel section only), is 19,600 in4.The area, A, is 73.3 in2.
TABLE 3.1
Dead Load Summary of Unfactored Shears and Moments for Interior and Exterior Girders
Girder Location VDC (kips) MDC (ft-kips) VDW (kips) MDW (ft-kips)
Interior 46.0 690.3 9.0 135.0Exterior 40.0 600.3 9.0 135.0
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The longitudinal stiffness parameter is
Kg = n(I + Aeg2) = (6.76)(19,600 in4 + (73.3 in3)(25.7 in)2)
= 459.77 in4
A Art. 4.6.2.2.1; A Eq. 4.6.2.2.1-1
The multiple presence factors have been included in the approximate equa-tions for the live distribution factors. These factors must be applied where the lever rule is specified.
A Tbl. 3.6.1.1.2-1; Comm. C3.6.1.1.2
The distribution factor for moments for interior girders of cross-section type (a) with one design lane loaded is
A Tbl. 4.6.2.2.2b-1 or Appendix A
DFM S SL
KLtsig
s= +
0 06
14 12
0 4 0 3
3.. .
0 1.
= +
0 06 1014
1060
459 70 4 0 3
. ,. .ft ft
ft774
12 60 7 5
4
3
0 1in
ft in( )( )
.
.
= 0.59
39.4 in2 = 19.7 in
N.A. steel39.4 in
eg=25.7 in
2 in4 in4 inN.A. deck
FIGURE 3.5Section for longitudinal stiffness parameter, Kg.
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where:S = spacing of beams (ft)L = span length (ft)ts = depth of concrete slab (in)
The distribution factor for moments for interior girders with two or more design lanes loaded is
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
= +
0 075 109 5
1060
4590 6 0 2
..
. .ft ft
ft,,
.
.774
12 60 7 5
4
3
0 1in
ft in( )( )
= 0.82 [controls]
The distribution factor for shear for interior girders of cross-section type (a) with one design lane loaded is
A Tbl. 4.6.2.2.3a-1 or Appendix C
DFV S ftsi = + = + =0 36
250 36 10
250 76. . .
The distribution factor for shear for interior girders with two or more design lanes loaded is
DFV S S ft ftmi = + −
= + − 0 212 35
0 2 1012
1035
2
. .
=2
0 952. [controls]
The distribution factors for exterior girders of typical cross section (a).
A Tbl. 4.6.2.2.1-1, 4.6.2.2.2d-1
Use the lever rule to find the distribution factor for moments for exterior girders with one design lane loaded. See Figure 3.6.
A Art. 3.6.1.3.1
ΣM@hinge = 0
02
9 52
3 5 10= ( ) + ( ) − ( )P ft P ft R ft. .
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R = 0.65 P
R = 0.65
The distribution factor for moments for exterior girders with one design lane loaded is
A Tbl. 3.6.1.1.2-1
DFMse = (R)(multiple presence factor for one lane loaded, m)
= (0.65)(1.2) = 0.78 [controls]
The distribution factor for moments for exterior girders with two or more design lanes loaded, g, is
A Tbl. 4.6.2.2.2d-1 or Appendix B
g = eginterior
where:ginterior = distribution factor for interior girder = DFMmi
g = DFMme = e(DFMmi)
The distance from the center of the exterior web of exterior beam to the interior edge of the barrier, de, is 1.5 ft.
A Tbl. 4.6.2.2.2d-1
10 ftde = 1.5 ft
R
1.5 ft 2 ft 6 ft 3.5 ft
hingeAssume
P2
P2
FIGURE 3.6Lever rule for the distribution factor for moments for exterior girder.
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The correction factor for distribution is
e d fte= + = + =0 779 1
0 77 1 59 1
0 935..
. ..
.
The distribution factor for moments for exterior girders with two or more design lanes loaded is
DFMme = e(DFMmi) = (0.935)(0.826) = 0.77
Use the lever rule to find the distribution factor for shear for exterior girder with one design lane loaded.
A Tbl. 4.6.2.2.3b-1 or Appendix D
This is the same as DFMse for exterior girders with one design lane loaded.
DFVse = DFMse = 0.78 [controls]
Find the distribution factor for shear for exterior girders with two or more design lanes loaded using g:
g = (e)(gint)
g = DFVme = (e)(DFVmi)
where:gint = distribution factor for interior girder = DFVmi
The correction factor for distribution is
A Tbl. 4.6.2.2.3b-1 or Appendix D
e d fte= + = + =0 610
0 6 1 510
0 75. . . .
The distribution factor for shear for exterior girders with two or more design lanes loaded is
DFVme = (0.75)(0.952) = 0.714
The HL-93 live load is made of the design lane load plus either the design truck (HS-20) or the design tandem load (whichever is larger).
A Art. 3.6.1.2
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Find the design truck (HS-20) moment due to live load. Please see Figure 3.7.
ΣM@B = 0
The reaction at A is
RA(60 ft) = (32 kips)(30 ft + 14 ft) + (32 kips)(30 ft) + (8 kips)(30 ft – 14 ft)
RA = 41.6 kips
The design truck moment due to live load is
Mtr = (41.6 kips)(30 ft) – (32 kips)(14 ft) = 800 ft-kips [controls]
Find the design tandem moment. See Figure 3.8.
ΣM@B = 0
The reaction at A is
RA(60 ft) = (25 kips)(30 ft) + (25 kips)(26 ft)
RA = 23.33 kips
C
8 kips32 kips32 kips14 ft 14 ft
30 ft30 ft
A
RA
B
FIGURE 3.7Design truck (HS-20) load moment at midspan.
C
25 kips25 kips
26 ft4 ft
30 ft30 ft
A
RA R2
B
FIGURE 3.8Design tandem load moment at midspan.
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The design tandem moment is
Mtandem = (23.33 kips)(30 ft) = 699.9 ft-kips (700 ft-kips)
Find the design lane moment. Please see Figure 3.9.
M M wLkips
ftft
fc lane= = =
( )
=2
2
8
0 64 60
8288
.tt-kips
The dynamic load allowance, IM, is 33% (applied to design truck or design tandem only, not to the design lane load).
A Art. 3.6.2.1
The unfactored total live load moment per lane is
MLL+IM = Mtr(1 + IM) + Mln
= (800 ft-kips)(1 + 0.33) + 288 ft-kips = 1352 ft-kips per lane
Find the design truck shear. See Figure 3.10.
ΣM@B = 0
32 kips 32 kips 8 kips
60 ft
14 ft14 ft
A
R R2
B
FIGURE 3.10Design truck (HS-20) shear at support.
C
30 ft
0.64 kips/ft
30 ft
A
R
B
FIGURE 3.9Design lane load moment.
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The reaction at A is
RA(60 ft) = (32 kips)(60 ft) + (32 kips)(46 ft) + (8 kips)(32 ft)
RA = 60.8 kips
The design truck shear, Vtr, is 60.8 kips [controls].Find the design tandem shear. See Figure 3.11.
ΣM@B = 0
The reaction at A is
RA(60 ft) = (25 kips)(60 ft) + (25 kips)(56 ft)
RA = 48.33 kips
The design tandem shear, Vtandem, is 48.33 kips.Find the design lane shear. See Figure 3.12.
V wLkips
ftft
kipsln
..= =
( )
=2
0 64 60
219 2
25 kips
60 ft
A
R R2
B
25 kips
4 ft
FIGURE 3.11Design tandem load shear at support.
0.64 kips/ft
60 ft
A
R
B
FIGURE 3.12Design lane load shear.
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The dynamic load allowance, IM, is 33% (applied to design truck or design tandem only, not to the design lane load).
A Art. 3.6.2.1The unfactored total live load shear is
VLL+IM = Vtr(1 + IM) + Vln
= (60.8 kips)(1 + 0.33) + 19.2 kips = 100.1 kips
Please see Table 3.2.Total live load moment and shear for interior girders
MLL+IM = 1108.6 ft-kips
VLL+IM = 95.3 kips per girder
Total live load moment and shear for exterior girders
MLL+IM = 1054.6 ft-kips
VLL+IM = 78.1 kips
Step 4: Check the Strength I Limit State.
A Arts. 3.3.2; 3.4; Tbls. 3.4.1-1, 3.4.1-2
U = 1.25(DC) + 1.5(DW) + 1.75(LL + IM)
The factored moment for interior girders is
Mu = 1.25 MDC + 1.5 MDW + 1.75 MLL+IM
= (1.25)(690.3 ft-kips) + (1.5)(135.0 ft-kips) + (1.75)(1108.6 ft-kips)
= 3005.4 ft-kips
TABLE 3.2 (A Art. 4.6.2.2.2)
Summary of Live Load Effects*
Girder Location
No. of Lanes
Loaded
Unfactored MLL+IM
(ft-kips per Lane) DFM
Unfactored MLL+IM
(ft-kips per Girder)
Unfactored VLL+IM
(kips per Lane) DFV
Unfactored VLL+IM (kips per Girder)
Interior 1 1352.0 0.596 100.1 0.76 —2 or more 1352.0 0.82 1108.6 100.1 0.952 95.3
Exterior 1 1352.0 0.78 1054.6 100.1 0.78 78.12 or more 1352.0 0.77 100.1 0.714 —
* The live load moments and shears per girder are determined by applying the distribution fac-tors, DF, to those per lane.
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The factored shear for interior girders is
Vu = 1.25 VDC + 1.5 VDW + 1.75 VLL+IM
= (1.25)(46 kips) + (1.5)(9.0 kips) + (1.75)(95.3 kips)
= 237.8 kips (238.0 kips)
The factored moment for exterior girders is
Mu = (1.25)(600.3 ft-kips) + (1.5)(135 ft-kips) + (1.75)(1054.6 ft-kips)
= 2798.4 ft-kips (2800 ft-kips)
The factored shear for exterior girders is
Vu = (1.25)(40 kips) + (1.5)(9.0 kips) + (1.75)(78.1 kips)
= 200.2 kips (200 kips)
Check the Flexural Resistance.
Find the flange stress, fbu, for noncomposite sections with continuously braced flanges in tension or compression. The required flange stress without the flange lateral bending, fbu, must satisfy the following.
A Art. 6.10.8.1.3
fbu ≤ ΦfRhFyf
A Eq. 6.10.8.1.3-1
Rh hybrid factor 1.0 (for rolled shapes)A Art. 6.10.1.10.1
Φf resistance factor for flexure 1.0A Art. 6.5.4.2
Fyf minimum yield strength of a flange 50 ksi
The flange stress for interior girders is
f MS
f ips inft
inbuu
x= =
( )
3005 4 12
992 3
. t-k
= 36.4 kips/in2 < ΦfRhFyf = 50 kips/in2 [OK]
Check the shear resistance in the web due to the factored loads, Vu, where:
A Arts. 6.10.9.1, 6.10.9.2
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Vp = plastic shear forceC = ratio of the shear–buckling resistance to the shear yield strength
A Art. 6.10.9.3.2 Vu ≤ ΦVn and Vn = CVp
Shear in the web, Vu, must satisfy the following.
Vu ≤ ΦvVn
A Eq. 6.10.9.1-1Nominal shear resistance,
A Eq. 6.10.9.2-1
Vn = CVp
Fyw specified minimum yield strength of a web 50 ksitw web thickness 0.75 inE modulus of elasticity of steel 29,000 ksiΦv resistance factor for shear 1.00
A Art. 6.5.4.2
D depth of steel beam 39.4 in
The shear yielding of the web isA Eq. 6.10.9.2-2
Vp = 0.58 FywDtw
= (0.58)(50 ksi)(39.4 in)(0.75 in)
= 856.95 kips (857 kips)
Find the ratio of the shear buckling resistance to the shear yield strength, C.C is determined from any of these AASHTO equations: Eq. 6.10.9.3.2-4,
6.10.9.3.2-5, 6.10.9.3.2-6.A Art. 6.10.9.3.2
Try AASHTO Eq. 6.10.9.3.2-4.
If Dt
EkFw yw
≤ 1 12. , then C = 1.0.
The shear buckling coefficient, k, is 5 for unstiffened web panels.A Com. 6.10.9.2
233Practice Problems
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Dt
ininw
= =39 40 75
52 5..
.
1 12 1 1229 000 5
5060 3. .
,.Ek
Fksi
ksiDtyw w
= ( )( ) = > = 552 5. [OK]
Therefore, C is 1.0.The nominal shear resistance for unstiffened webs is
A Eq. 6.10.9.2-1
Vn = CVp = (1.0)(857 kips) = 857 kips [OK]
The shear resistance for interior girders is greater than the factored shear for interior girder.
ΦvVn = (1.0)(857 kips) = 857 kips > Vu = 238 kips [OK]A Art. 6.10.9
Step 5: Check the Fatigue Limit State
Details shall be investigated for fatigue as specified in AASHTO Art. 6.6.1. The fatigue load combination in AASHTO Tbl. 3.4.1-1 and the fatigue live load specified in AASHTO Art. 3.6.1.4 shall apply.
A Art. 6.10.5.1
For the load induced fatigue, each detail must satisfy the following:A Art. 6.6.1.2, A Eq. 6.6.1.2.2-1
γ(Δf) ≤ (ΔF)n
γ load factor(Δf) live load stress range due to fatigue load ksi(ΔF)n nominal fatigue resistance as specified in Art. 6.6.1.2.5 ksi
The fatigue load is one design truck with a constant spacing of 30 ft between 32 kip axles. Find the moment due to fatigue load.
A Art. 3.6.1.4
Find the center of gravity of design truck, x, from the left 32 kip axle
xkips ft kips ft ft
kips= ( )( ) + ( ) +( )
+32 30 8 30 14
32 322 8kips kips+
= 18.22 ft from left 32 kip axle
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Please see Figures 3.13 and 3.14.
ΣM@B = 0
The reaction at A is
RA(60 ft) = (32 kips)(54.11 ft) + (32 kips)(24.11 ft) + (8 kips)(10.11 ft)
RA = 43.07 kips
The maximum moment at point E,
ME = (43.07 kips)(35.89 ft) – (32 kips)(30 ft) = 585.8 ft-kips [controls]
The maximum moment due to fatigue load is 585.8 ft-kips.
8 kips32 kips 32 kips
30 ft 14 ft
5.89 ft5.89 ft
c.g. ofdesign truck
x = 18.22 ft30 ft – 18.22 ft
2 = 5.89 ft
D
FIGURE 3.13Center of gravity of design truck loading (HS-20).
30 ft
32 kips 32 kips 8 kips5.89 ft 30 ft 14 ft 10.11 ft
24.11 ft24.11 ft 5.89 ft
A
RA
B
30 ft
E
CL
FIGURE 3.14Fatigue load position for maximum moment.
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Find the shear due to the Fatigue Load, Vfat. See Figure 3.15.
ΣM@B = 0
The reaction at A is
R kips kips kips ftftA = ( ) +
+ ( )
32 322
8 1660
= 50.13 kips
Vfat = 50.13 kips
The dynamic load allowance, IM, is 15%.A Tbl. 3.6.2.1-1
Find the Load Distribution for Fatigue, DFMfat.The distribution factor for one traffic lane shall be used. For single lane,
approximate distribution factors in AASHTO Art. 4.6.2.2 are used. The force effects shall be divided by 1.20.
A Art. 3.6.1.4.3b, 3.6.1.1.2
The distribution factor for fatigue moments in interior girders with one lane is
DFM DFMfat,int
int
...
.= = =1 2
0 591 2
0 49
The distribution factor for the fatigue moments in exterior girders with one loaded lane is
DFM DFMfat ext
ext, .
..
.= = =1 2
0 781 2
0 65
30 ft
32 kips 32 kips 8 kips30 ft 14 ft 16 ft
A
RA
B
30 ft
C
FIGURE 3.15Fatigue load position for maximum shear.
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The unfactored distributed moment for interior girders is
Mfat,int = Mfatigue(DFMfat,int)(1 + IM)
= (585.8 ft-kips)(0.49)(1 + 0.15) = 330.1 ft-kips (330 ft-kips)
The unfactored distributed moment for exterior girders is
Mfat,ext = Mfatigue(DFMfat,ext)(1 + IM)
= (585.8 ft-kips)(0.65)(1 + 0.15) = 437.9 ft-kips (438 ft-kips)
Live load stress due to fatigue load for interior girders is
∆f MS
f s inftfat
xint
,int.
( ) = =
330 0 12t-kip
99923 993
2
inkips in= . /
Live load stress due to fatigue load for exterior girders is
∆f MS
ips inft
extfat ext
x( ) = =
,
438 12
99
ft-k
225 303
2
inkips in= . /
For load-induced fatigue, each detail shall satisfy:A Art. 6.6.1.2.2
γ(Δf) ≤ (ΔF)n
Find the nominal fatigue resistance, (ΔF)n, for Fatigue II load combination for finite life.
A Art. 6.6.1.2.5, A Eq. 6.6.1.2.5-2
∆F ANn( ) =
13
The girder is in Detail Category A, because it is a plain rolled member.A Tbl. 6.6.1.2.3-1
A = constant taken for Detail Category A 250 × 108 kips/in2
A Tbl. 6.6.1.2.5-1
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n = number of stress range cycles 1.0 (for simple span girdersper truck passage with span greater than 40 ft)
A Tbl. 6.6.1.2.5-2
p = the fraction of truck traffic in a 0.80 (because NL is 3 lanes)single lane
A Tbl. 3.6.1.4.2-1
(ΔF)TH = the constant amplitude threshold 24.0 kips/in2
for Detail Category A
A Tbl. 6.6.1.2.5-3
ADTTSL = the number of trucks per day in a single lane over the design life
ADTT = the number of trucks per day in one direction over the design life
ADTTSL = (p)(ADTT) = (0.80)(250) = 200 trucks daily
A Art. 3.6.1.4.2
The number of cycles of stress range N is
N = (365 days)(75 years)n(ADTT)SL = (365 days)(75 years)(1.0)(200 trucks)
= 5,475,000 cycles
A Eq. 6.6.1.2.5-3The nominal fatigue resistance is
A Eq. 6.6.1.2.5-2
∆F A
N
kipsin
cycn( ) =
=×1
38
2250 10
5 475 000, , lleskips in
=
13
216 6. /
Find the Load Factor, γ, and confirm that the following are satisfied.
A Eq. 6.6.1.2.2-1 γ(Δf) ≤ (ΔF)n
γ = 0.75 for Fatigue II Limit State
A Tbl. 3.4.1-1
The factored live load stress due to the fatigue load for interior girders is
γ(Δfint) = (0.75)(3.99 ksi) = 2.99 ksi < 16.6 ksi [OK]
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The factored live load stress due to the fatigue load for exterior girders is
γ(Δfext) = (0.75)(5.30 ksi) = 3.98 ksi < 16.6 ksi [OK]
A Art. 6.10.5.3
Special Fatigue Requirement for Webs
Find the shear in the web due to unfactored permanent load plus factored fatigue shear load, Vu, for interior and exterior girders, satisfying the follow-ing provision to control web-buckling and elastic flexing of the web. Vcr is the shear buckling resistance (Eq. 6.10.9.3.3-1). The use of the Fatigue I load pro-visions for the Fatigue II load combination will be considered conservative.
A Eq. 6.10.9.3.3-1
The distribution factor for fatigue shear in interior girders with one design lane loaded is
DFV DFVmfat,int
int ..
.= = =0 761 2
0 63
The distribution factor for fatigue shear in exterior girders with one design lane loaded is
DFV DFVmfat ext
ext,
..
.= = =0 781 2
0 65
Unfactored fatigue shear load per interior girder is
(DFVfat,int)(Vfat) = (0.63)(50.13 kips) = 31.58 kips
Unfactored fatigue shear load per exterior girder is
(DFVfat,ext)(Vfat) = (0.65)(50.13 kips) = 32.58 kips
Vu ≤ Vcr
A Eq. 6.10.5.3-1 Vu = VDL + γ(Vfat)(1 + IM)
The shear buckling resistance for unstiffened webs, Vcr, for W40 × 249 with Fy equal to 50 ksi is
A Art. 6.10.9.2; Eq. 6.10.9.2-1 Vcr = CVp
C = 1.0A Eq. 6.10.9.3.2-4
239Practice Problems
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Vp = 0.58 FywDtw = 857 kips
A Eq. 6.10.9.2-2
Vcr = CVp = (1.0)(857 kips) = 857 kips
The distortion-induced fatigue for interior girders is
VDL = VDC + VDW
Vu = VDL + γVfat(1 + IM) = (46 kips + 9 kips) + (0.75)(31.58 kips)(1.15)
= 82.23 kips < Vcr = 857 kips[OK]
The distortion induced fatigue for exterior girders is
Vu = VDL + γVfat(1 + IM) = (40 kips + 9 kips) + (0.75)(32.58 kips)(1.15)
= 77.1 kips < Vcr = 857 kips[OK]
Step 6: Check the Service Limit States
Deflection Limit
A Art. 2.5.2.6
The maximum deflection for vehicular load, δmax, where L is the span, is
A Art. 2.5.2.6.2
δmax = L800
Deflection will be the larger of
A Art. 3.6.1.3.2
1. Deflection resulting from the design truck (HS-20) alone
2. Deflection resulting from 25% of design truck plus the design lane load
Find the deflection resulting from the design truck alone (see Figure 3.16)
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The deflections at midspan, δ1, δ2, δ3, are
δ11 2 2
22 2
6= − −( )P b x
EILL b x
=( ) +( )( )
( )
32 30 14 30 12
6 2
2
kips ft ft ft inft
99 000 19 600 60 12
60
4, ,ksi in ft inft
f
( )( )( )
tt inft
ft ft inft
12 30 14 122
− +( )
−
2 2
30 12ft inft
= 0.273 in
δ22
33
3
48
32 60 12
48= =
( )( )
(P L
EI
kips ft inft
))( )( )29 000 19 600 4, ,ksi in
= 0.438 in
P2 = 32 kipsP1 = 32 kips
1
P3 = 8 kipsx = 30 ft 16 ft14 ft14 ft
A BC
b1
b2a
2 3
30 ft 30 ft
L
FIGURE 3.16Design truck loading for maximum deflection at midspan.
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δ33 1 2
12 2
6= − −( )P b x
EILL b x
=( ) +( )( )
( )
8 30 14 30 12
6 29
2
kips ft ft ft inft
,, ,000 19 600 60 12
60
4ksi in ft inft
ft
( )( )( )
112 30 14 122
inft
ft ft inft
− +( )
−
2 2
30 12ft inft
= 0.068 in
δLL = δ1 + δ2 + δ3 = 0.273 in + 0.438 in + 0.068 in = 0.779 in per lane
m multiple presence factor for three loaded lanes 0.85
A Tbl. 3.6.1.1.2-1
NL number of lanes 3
Ng number of girders 5
The dynamic load allowance, IM, is 33%.
A Art. 3.6.2.1
Distribution factor for deflection is equal to the number of lanes, NL, divided by the number of beams, Ng.
A Comm. 2.5.2.6.2
DF m NN
L
g= ( )
A Tbl. 3.6.2.1-1
DF = ( )
=0 85 35
0 51. .
IM = 33%
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The distributed midspan deflection per girder due to the design truck alone is
δLL+IM = (0.779 in per lane)(0.51)(1 + 0.33) = 0.53 in per girder [controls]
A Art. 3.6.1.3.2
Find the deflection resulting from 25% of design truck plus the lane load.
A Art. 3.6.1.3.2
The deflection at midspan due to the lane load is (see Figure 3.17)
δln
.= =
( )
5384
5 0 64124wL
EI
kipsft
ftin
( )
( )( )60 12
384 29 000 19 60
4
ft inft
ksi, , 000 334in
in( ) = . per lane
The midspan deflection per lane is
δLL25%+ln = 25% of deflection due to design truck + δln
= (0.25)(δLL) + δln
= (0.25)(0.779 in) + 0.33 in = 0.525 in per lane [controls]
The distributed midspan deflection per girder is
δLL+IM = (δLL25%+ln)(DF)(1 + IM) = (0.525 in per lane)(0.51)(1 + 0.33)
= 0.356 in per girder
Confirm that δLL+IM < δmax.
C
0.64 kips/ft = Lane load
60 ft
FIGURE 3.17Design lane loading for maximum deflection at midspan.
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δmax .= =( )
=L
ft inft in
800
60 12
8000 90
δLL+IM = 0.525 in < δmax = 0.90 in [OK]A Art. 6.10.4.2
Permanent deformations
For inelastic deformation limitations the Service II Limit State load com-bination shall apply and both steel flanges of noncomposite sections shall satisfy the following.
Flexure (to prevent objectionable permanent deflections)
A Art. 6.10.4.2.2
Both steel flanges of noncomposite sections shall satisfy the following, where ff is the flange stress due to Service II loads.
A Eq. 6.10.4.2.2-3
f f R Ffl
h yf+ ≤2
0 80.
fl flange lateral bending stress due to 0 (for continually braced flanges) Service II loads Rh hybrid factor 1.0 (for rolled shapes)
A Art. 6.10.1.10.1
Service II Limit State: U = 1.0 DC + 1.0 DW + 1.3(LL + IM)
A Tbl. 3.4.1-1
The factored design moment for interior girders is
Mu = 1.0 MDC + 1.0 MDW + 1.3 MLL+IM
= (1.0)(690.3 ft-kips) + (1.0)(135 ft-kips + (1.3)(1108.6 ft-kips))
= 2266.5 ft-kips
The flange stress due to Service II loads for interior girders is
f MS
f s inft
infu
x= =
=
2266 5 12
992273
..
t-kip44 2kips in/
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For steel flanges of noncomposite sections in interior girders, check,
A Art. 6.10.4.2.2, A Eq. 6.10.4.2.2-3
f f R Ffl
h yf+ ≤2
0 80.
27 4 02
27 4 0 8 1 0 50 40. . . .ksi ksi ksi ksi+ = < ( )( )( ) = [OK]
The factored design moment for exterior girders is
Mu = (1.0)(600.3 ft-kips) + (1.0)(135 ft-kips) + (1.3)(1054.6 ft-kips)
= 2106.8 ft-kips
The flange stress due to Service II loads for exterior girders is
f MS
s inft
infu
x= =
( )
=
2106 8 12
99223
. ft-kip55 5. ksi
For steel flanges of noncomposite section in exterior girders, check,
f f R Ffl
h yf+ ≤2
0 80.
25 5 02 25 5 0 8 1 0 50 40. . . .ksi ksi ksi ksi+ = < =( )( )( ) [OK]
Practice Problem 2: 161 ft Steel I-Beam Bridge with Concrete Slab
Situation
L span length 161 ftLL live load model HL-93S beam spacing 13 ftts slab thickness (including 0.5 in integral wearing surface) 10.5 in self-weight of steel girder 335 lbf/ft
245Practice Problems
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FWS future wearing surface 25 lbf/ft2
weight of stiffeners and bracings 10% of girder weight dead load of the curb/parapet 0.505 kips/ft self-weight of stay-in-place (SIP) forms 7 lbf/ft2
f′c concrete strength 5.0 kips/in2
wc concrete unit weight 145 lbf/ft3
The steel girders satisfy the provisions of AASHTO Art. 6.10.1 and 6.10.2 for the cross-section properties’ proportion limits. See Figure 3.18.
Solution
Steel Girder Section Review. See Figure 3.19
A area of girder 106.0 in2
yb distance from bottom fiber to centroid of section 30.88 inyt distance from top fiber to centroid of section 40.12 inI moment of inertia about steel section centroid 99,734.0 in4
Please see Figure 3.20.
161 ft
(a) Elevation
39 ft
(b) Cross section
13 ft 13 ft
2 in Haunch
13 ft
4.25 ft4.25 ft
2.5 ft 2.5 ft1.75 ft
1.75 ft 1.75 ft44 ft roadway
1.75 ft
FIGURE 3.18Steel I-beam with concrete slab.
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Step 1: Determine Dead Load Shears and Moments for Interior and Exterior Girders
Dead Loads DC: Components and Attachments
1.1 Interior Beam
1.1.a Noncomposite Dead Loads, DC1 The steel beam weight plus 10% for stiff-eners and bracings is,
DC kipsftbeam =
( )0 335 1 10. .
DC kipsftbeam = 0 369.
22 in
c.g.
2.0 in
67.75 in0.5 in
40.1
2 in
30.8
8 in
d = 71.0 in
1.25 in
22 in
FIGURE 3.19I-beam properties.
30.88 in
9.5 in
2 in haunch
c.g.Steel section
40.12 in
Slab of deck c.g.
eg
4.75 in
13 ft
FIGURE 3.20Cross-section properties for shears and moments due to dead loads.
247Practice Problems
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The slab dead load is
DC in ftin
ft kipsfslab = ( )
( )10 5 112
13 0 15. .tt3
DC kipsftslab = 1 70.
The concrete haunch dead load is
DC in in ftin
kiphaunch = ( )( )
2 22 1144
0 152
2 . ssft
DC kipsfthaunch = 0 046.
The stay-in-place (SIP) forms dead load is
DC kipsft
ftgirdersSIP =
( )
0 007 44 0 1
42. .
DC kipsftSIP = 0 077.
DC DC DC DC DCbeam slab haunch SIP1 = + + +
DC kipsft
kipsft
kipsft1 0 369 1 70 0 046 0 07= + + +. . . . 77 kips
ft
DC kipsft1 2 19= .
Shear
V wLkips
ftft
DC1 2
2 19 161
2= =
( ).
V kipsDC1 176 3= .
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Moment
M wLkips
ftft
DC1
22
8
2 19 161
8= =
( ).
M ki tDC1 7 095 9= , . p-f
1.1.b Composite Dead Loads Due to Curb/Parapet, DC2
DC kipsft girders2 0 505 2 1
4=
( )
.
DC kipsft2 0 253= .
Shear
V wLkips
ftft
DC2 2
0 253 161
2= =
( ).
V kipsDC2 20 4= .
Moment
M wLkips
ftft
DC2
22
8
0 253 161
8= =
( ).
M kiDC2 819 8= . p-ft
1.1.c Total Dead Load Effects Due to DC = DC1 + DC2
Shear
V V VDC DC DC= +1 2
V kips kipsDC = +176 3 20 4. .
V kipsDC = 196 7.
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Moment
M M MDC DC DC= +1 2
M ki t ki tDC = +7 095 9 819 8. . .p-f p-f
M kiDC = 7 915 7, . p-ft
1.1.d Dead Loads, DW; Wearing Surfaces
For future wearing surface of 25 lbf/ft2
DW lbfft
kiplbf
ftg
=
( )25 11000
44 142 iirders
DW kipsft
= 0 275.
Shear
V wLkips
ftft
DW = =
( )
2
0 275 161
2
.
V kipsDW = 22 14.
Moment
M wLkips
ftft
DW = =
( )2
2
8
0 275 161
8
.
M ki tDW = 891 0. p-f
Please see Table 3.3 for summary in interior beams.
1.2 Exterior Beams
The exterior beam has a slab overhang of 4.25 ft. Therefore, the interior beam dead loads control for design.
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Step 2: Determine Live Load Shears and Moments Using the AASHTO HL-93 Load.
2.1 Determine Live Load Shears Due to HL-93 Loads
Please see Figure 3.21.
ΣMB = 0
(32 kips)(161 ft) + (32 kips)(161 ft – 14 ft) + (8 kips)(161 ft – 28 ft) – R(161 ft) = 0
R = 67.8 kips
Vtruck = 67.8 kips [controls]
Please see Figure 3.22.
TABLE 3.3
Summary of Dead Load Shears and Moments in Interior Beams
DC
DC1 DC2 DC1 + DC2 DW
Shear, kips 176.3 20.4 196.7 22.1Moment, kip-ft 7095.9 819.8 7915.7 891.0
161 ft
14 ft14 ft
32 kips 32 kips 8 kips
R
A B
FIGURE 3.21Design truck load (HS-20) position for maximum shear.
161 ft B
4 ft
25 kips 25 kips
R
A
FIGURE 3.22Design tandem load position for maximum shear.
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ΣMB = 0
(25 kips)(161 ft) + (25 kips)(161 ft – 4 ft) – R(161 ft) = 0
R = 49.4 kips
Vtandem = 49.4 kips
Please see Figure 3.23.
ΣMB = 0
(0.640 kips/ft)(161 ft)(1/2) – R(161 ft) = 0
R = 67.8 kips
Vlane = 51.5 kips
2.2 Determine Live Load Moments Due to HL-93 Loads
Please see Figure 3.24 for design truck moment.
ΣMB = 0
(32 kips)(80.5 ft + 14 ft) + (32 kips)(80.5 ft) + (8 kips)(80.5 ft – 14 ft) – R(161 ft) = 0
R = 38.09 kips
161 ft
B
0.640 kips/ft
R
A
FIGURE 3.23Design lane load position for maximum shear.
161 ft
80.5 ft
B14 ft
32 kips 8 kips
14 ft
32 kips
R
A
80.5 ft
FIGURE 3.24Design truck load (HS-20) position for maximum moment.
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Mtruck = (38.09 kips)(80.5 ft) – (32 kips)(14 ft)
Mtruck = 2618.2 kip-ft [controls]
Please see Figure 3.25 for design tandem load moment.
ΣMB = 0
(25 kips)(80.5 ft) + (25 kips)(80.5 ft – 4 ft) – R(161 ft) = 0
R = 24.38 kips
Mtandem = (24.38 kips)(80.5 ft)
Mtandem = 1962.6 kip-ft
Please see Figure 3.26 for design lane moment.
M wLkips
ftft
lane = =
( )2
2
8
0 640 161
8
.
M k tlane = 2073 7. ip-f
Please see Table 3.4, summary of live load shears and moments.
80.5 ft
B4 ft
25 kips 25 kips
76.5 ft
R
A
80.5 ft
FIGURE 3.25Design tandem load position for maximum moment.
161 ft
B
0.640 kips/ft
A
FIGURE 3.26Design lane load position for maximum moment.
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Step 3: Determine Live Load Distribution Factors
A Art. 4.6.2.2.2; Tbl. 4.6.2.2.1-1
3.1 For Moment in Interior Beam, DFM
A Art. 4.6.2.2.2bLongitudinal stiffness parameter Kg is,
A Eq. 4.6.2.2.1-1
K n I Aeg g= +( )2
where:
n EE
s
c=
Es = modulus of elasticity of steel = 29 x 103 kips/in2
Ec = modulus of elasticity of deck concrete materialA Comm., Eq. 5.4.2.4-1
E fc c= ′1820 for wc = 0.145 kips/ft3
= 1820 5 2kipsin
= 4069 6 2. kipsin
eg = distance between the centers of gravity of the beam and deckeg = 40.12 in + 2 in + 4.75 ineg = 46.87 in
n = ××
=29 104 0696 10
7 133
3..
Use n = 7.0.
TABLE 3.4
Summary of Live Load Shears and Moments
Truck Load Effect Tandem Load Effect Lane Load Effect
Shear, kips 67.8 49.4 51.5Moment, kip-ft 2618.2 1962.6 2073.7
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A = area of basic beam sectionA = 106.0 in2 (given)I = moment of inertia of basic beam sectionI = 99,734.0 in4 (given)Kg = 7.0 [99734.0 in4 + (106 in2)(46.87 in)2] = 2.33 x 106 in4
Distribution of live loads per lane for moment in interior beams with cross-section type (a).
A Tbl. 4.6.2.2.1-1, 4.6.2.2.2b-1 or Appendix A
One design lane loaded:
m DFM DFM S SLsi si( )( ) = = +
0 0614
0 4 0
.. .. .3
3
0 1
12KLt
g
s
where:m = multiple presence factors have been included in the approximate
equations for distribution factors. They are applied where the lever rule is used.
A Art. 3.6.1.1.2; Tbl. 3.6.1.1.2-1
si = single lane loaded, interior girderDFM = moment distribution factorS = spacing of beams (ft)L = span length of beam (ft)ts = depth of concrete slab (in)
DFM ft ftftsi = +
0 06 1314
13161
0 4 0
.. .33 6 4
3
0 1
2 33 1012 161 9 5
..
.
×
( )( )
inft in
= 0.532
Two or more lanes loaded:
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
where:mi = multiple lanes loaded, interior girder
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DFM ft ftftmi = +
0 075 139 5
13161
0 6
..
. 00 2 6 4
3
0 1
2 33 1012 161 9 5
. .
..
×
( )( )
inft in
= 0.83
3.2 Distribution of Live Load per Lane for Moment in Exterior Beams with Cross-Section Type (a)
A Tbl. 4.6.2.2.2d-1 or Appendix B
One design lane loaded:
Use lever rule. See Figure 3.27.A Fig. C4.6.2.2.1-1
ΣMA = 0
(P/2)(13.5 ft/2) + (P/2)(7.5 ft) – R(13 ft) = 0
R = 0.808 P
DFMse = (m)(0.808)
R = (1.2)(0.808) = 0.970
where:se = single lane loaded, exterior girderm = multiple presence factor for one lane loaded = 1.2
13.0'9.25 ft R
7.5'
Assumed hinge
A5.5'
6.0'0.5'
1.75'
2.0'
2P
2.5'
2P
FIGURE 3.27Lever rule for the distribution factor for moments for exterior girder.
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Two or More Design Lanes Loaded:
g = (e)(ginterior) orDFMme = (e)DFMmi
e de= +0 779 1
..
where:g = DFM = distribution factors for momentginterior = DFMmi = distribution factor for interior girder = 0.83de = distance from the centerline of the exterior beam web to the inte-
rior edge of curb (ft) = 2.5 ft
e ft ft= + =0 77 2 59 1
1 045. ..
.
DFMme = (1.045 ft)(0.83)
= 0.867 or
DFMme = g = 0.867
3.3 Distribution of Live Load per Lane for Shear in Interior Beams.
A Tbl. 4.6.2.2.3a-1 or Appendix C
One design lane loaded:
DFV Ssi = +0 36
25 0.
.
= +0 36 1325 0
..ft
= 0.88
where:DFV = shear distribution factor
Two or more lanes loaded:
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DFV S Smi = + −
0 212 35
2 0
..
= + −
0 2 13
121335
2 0
..ft ft
= 1.145
3.4 Distribution of Live Load per Lane for Shear in Exterior Beams
A Tbl. 4.6.2.2.3b-1 or Appendix D
One design lane loaded:Use lever rule (same as the distribution factors for moment)
DFV DFMse se=
= 0.970
Two or more design lanes loaded:
DFVse = (e)(ginterior)
ginterior = DFVmi
DFV e DFVme mi= ( )( )
where:
e de= +0 610
.
e ft= +0 6 2 510
. .
e = 0.85
DFVme = ( )( )0 85 1 145. .
= 0.973
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Please see Table 3.5.
Step 4: Unfactored Distributed Live Load per Beam with Impact
Dynamic load allowance, IM, is equal to
33% for design truck or tandem
0% for design lane
15% for Fatigue and Fracture Limit State
A Art. 3.6.2.1
4.1 Shear
V DFV V or V IM VLL IM truck dem lane+ = ( ) +( ) + tan 1
4.1.a Interior Beam
V kips kipsLL IM+ = ( ) ( )( ) + 1 145 67 8 1 33 51 5. . . .
V kipsLL IM+ = 162 2.
4.1.b Exterior Beam
V kips kipsLL IM+ = ( ) ( )( ) + 0 973 67 8 1 33 51 5. . . .
V kipsLL IM+ = 137 8.
TABLE 3.5
Summary of Live Load Distribution Factors
Moment Shear
Distribution Factor Equation
Interior Beam
Exterior Beam
Interior Beam
Exterior Beam
One Lane Loaded Approximate 0.532 — 0.88 —Lever Rule — 0.970 — 0.970
Two or More Lanes Loaded Approximate 0.830 0.867 1.145 —Lever Rule — — — 0.973
Controlling Value 0.830 0.970 1.145 0.973
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4.2 Moment
M DFM M or M IM MLL IM truck dem lane+ = ( ) +( ) + tan 1
4.2.a Interior Beam
M ki kLL IM+ = ( )( ) +0 83 2618 2 1 33 2073 7. . . .p-ft ip-ftt
M kLL IM+ = 4611 4. ip-ft
4.2.b Exterior Beam
M ki t kiLL IM+ = ( )( ) +0 97 2618 2 1 33 2073 7. . . .p-f p-ftt
M ki tLL IM+ = 5389 2. p-f
Please see Table 3.6.
Step 5: Factored Design Loads for Limit State Strength I
A Tbl. 3.4.1-1, 3.4.1-2
5.1 Shear, Vu
V V V Vu DC DW LL IM= + + +1 25 1 5 1 75. . .
5.1.a Interior Beam
V kips kips ku = ( ) + ( ) +1 25 196 7 1 5 22 1 1 75 162 2. . . . . . iips( )
V kipsu = 562 9.
TABLE 3.6
Summary of Unfactored Distributed Live Load Effects per Beam
Beams VLL+IM, kips MLL+IM, kip-ft
Interior 162.2 4611.4Exterior 137.8 5389.2
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5.1.b Exterior Beam
V kips kips ku = ( ) + ( ) +1 25 196 7 1 5 22 1 1 75 137 8. . . . . . iips( )
V kipsu = 520 2.
5.2 Moment, Mu
M M M Mu DC DW LL IM= + + +1 25 1 5 1 75. . .
5.2.a Interior Beam
M k t ki tu = ( ) + ( ) +1 25 7915 7 1 5 891 0 1 75. . . . .ip-f p-f 44611 4. kip-ft( )
M ku = 19301 0. ip-ft
5.2.b Exterior Beam
M ki t ki tu = ( ) + ( ) +1 25 7915 7 1 5 891 0 1 75. . . . .p-f p-f 55389 2. k tip-f( )
M ki tu = 20662 2. p-f
Step 6: Fatigue II Limit State for Finite Load-Induced Fatigue Life
A Arts. 3.6.1.4, 3.6.1.1.2, 3.6.2.1
6.1 Fatigue Load
Fatigue load shall be one design truck or axles, but with a constant spacing of 30 ft between the 32 kip axles. Please see Figure 3.28.
80.5 ft
B
14 ft30 ft32 kips32 kips 8 kips
RA
A
80.5 ft
FIGURE 3.28Fatigue load position for maximum moment at midspan.
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ΣMB = 0
(32 kips)(80.5 ft + 30 ft) + (32 kips)(80.5 ft) + (8 kips)(80.5 ft –14 ft) – RA(161.0 ft) = 0
RA = 41.27 kips
The maximum fatigue load moment, Mfat, is,
Mfat = (41.27 kips)(80.5 ft) –(32 kips)(30.0 ft)
Mfat = 2362.0 kip-ft
The multiple presence factors have been included in the approximate equations for distribution factors in AASHTO [4.6.2.2 and 4.6.2.3], both for single and multiple lanes loaded. Where the lever rule (a sketch is required to determine load distributions) is used, the multiple presence factors must be included. Therefore, for fatigue investigations in which fatigue truck is placed in a single lane, the factor 1.2 which has been included in the approxi-mate equations should be removed.
A Comm. 3.6.1.1.2
6.2 Fatigue Live Load Distribution Factors per Lane for Moment
The dynamic allowance for Fatigue Limit State, IM, is 15%.
A Tbl. 3.6.2.1-1
Determine distribution factors for moments, DFMfat,
6.2.a For Interior Beam
DFM DFMmfat
Isi= ( )
1
DFMfatI = ( )
0 532 11 2
..
DFMfatI = 0 44.
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6.2.b For Exterior Beam
DFM DFMmfat
Ese= ( )
1
DFMfatE = ( )
0 97 11 2
..
DFMfatE = 0 81.
6.3 Unfactored Distributed Fatigue Live Load Moment per Beam with Impact
M DFM M IMfat IM fat fat+ = ( )( ) +( )1
6.3.a Interior Beam
M DFM M IMfat IMI
fatI
fat+ = ( )( ) +( )1
M tfat IMI
+ = ( )( ) +( )0 44 2362 0 1 0 15. . .kip-f
M k tfat IMI
+ = 1195 2. ip-f
6.3.b Exterior Beam
M DFM M IMfat IME
fatE
fat+ = ( )( ) +( )1
M ki tfat IME
+ = ( )( ) +( )0 81 2362 0 1 0 15. . .p-f
M kfat IMI
+ = 2200 2. ip-ft
6.4 Factored Fatigue II Design Live Load Moment per Beam, MQ
A Tbl. 3.4.1-1
Q MLL IM= +0 75.
6.4.a Interior Beam
M MQ fatI
fat IMI
, .= ( )+0 75
M k tQ fatI
, . .= ( )0 75 1195 2 ip-f
M kQ fatI
, .= 896 4 ip-ft
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6.4.b Exterior Beam
M MQ fatE
fat IME
, .= ( )+0 75
M ki tQ fatI
, . .= ( )0 75 2200 2 p-f
M kiQ fatI
, .= 1650 2 p-ft
Practice Problem 3: Interior Prestressed Concrete I-Beam
Situation
The following design specifications apply to a bridge in western Pennsylvania.
L bridge span 80 ft
f′ci compressive strength of concrete at time of initial
prestress 5.5 ksi
f′cg compressive strength of concrete at 28 days for
prestressed I-beams 6.5 ksi
number of roadway prestressed 24 × 54 in I-beams 6 beams
S beam spacing 8 ft
ts roadway slab thickness 7.5 in
integral wearing surface of slab 0.50 in
clear roadway width 44 ft 6 in
f′cs compressive strength of roadway slab concrete
at 28 days 4.5 ksi
wFWS future wearing surface dead load 0.030 kips/ft2
ws = wC&P curb and parapet dead load 0.506 kips/ft
Aps area of prestressing steel (0.5 in diameter
low-relaxation strand; seven wire) 0.153 in2
Ep modulus of elasticity of prestressing steel 28,500 ksi
fpu ultimate stress of prestressing steel (stress relieved) 270 ksi
A composite deck and a standard curb and parapet are used. Assume no haunch for composite section properties and 150 lbf/ft3 concrete for all com-ponents. The bridge cross section is shown. See Figures 3.29, 3.30, and 3.31.
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Requirement
Review the interior bonded prestressed concrete I-beam for the Load Combination Limit State Strength I.
where:Qi = force effectη = load modifierγ = load factorηi = ηD ηR ηI = 1.0
Solution
Strength I Limit State: Q = 1.0(1.25 DC + 1.50 DW + 1.75(LL + IM)). Please see Table 3.7.
A Tbl. 3.4.1-1, A Art. 1.3.3, 1.3.4, 1.3.5, 1.3.2.1
80 ft
(a) Elevation
4 ft4 ft5 @ 8 ft = 40 ft
44 ft 6 in 1 ft 9 in
2 ft 8 in8 in slab (7.5 in structural)
(b) Cross section
FIGURE 3.29Prestressed concrete I-beam.
8 ft
24 in × 54 inI-beam
7.5 in slab + 0.5 in
8 ft
FIGURE 3.30Deck and I-beam.
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Step 1: Determine the Cross-Sectional Properties for a Typical Interior Beam
The properties of the basic beam section are given as follows:
Ag area of basic beam 816 in2
yb center of gravity of the basic beam from the bottom of the basic beam 25.31 inyt distance from the center of gravity of the basic beam to the top of the basic beam 28.69 inh basic beam depth 54.0 inIg moment of inertia of the basic beam section about centroidal axis, neglecting reinforcement 255,194 in4
TABLE 3.7
Load Modifiers
Strength Service Fatigue
Ductility, ηd 1.0 1.0 1.0
Redundancy, ηr 1.0 1.0 1.0
Importance, ηi 1.0 N/A N/A
η = ηDηRηI ≥ 0.95 1.0 1.0 1.0
3 in
1 ft 9 in
0.68 ft
10 in
1 ft 1 in
2 ft 9 in
7 in1 ft
2 in
FIGURE 3.31Curb and parapet.
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The section moduli for the extreme fiber of the noncomposite section are
SIy
inin
inncbg
b= = =255 194
25 3110 083
43,
.,
SI
h yin
in ininnct
g
b=
−=
−=255 194
54 0 25 318895
4,. .
33
The unit weight of concrete, wc, is 0.150 kips/ft3.
A Tbl. 3.5.1-1
The modulus of elasticity of concrete at transfer is
A Eq. 5.4.2.4-1
E w f kips
ftci c ci= ′ = ( )
33 000 33 000 0 1501 53, , ..
=
1 5
5 5 4496.
. ksi ksi
The modulus of elasticity for the prestressed I-beam concrete at 28 days is
E w f kipsftcg c cg= ′ = ( )
33 000 33 000 0 1501 5
3, , ..
=1 5
6 5 4888.
. ksi ksi
The modulus of elasticity of the roadway slab concrete at 28 days is
E w f kipsftcs c cs= ′ = ( )
33 000 33 000 0 1501 5
3, , ..
=1 5
4 5 4067.
. ksi ksi
Please see Figure 3.32.
bw web width 8 in
btop top flange width 18 in
The effective flange width for interior beams is
A Art. 4.6.2.6.1
bi = S = 8 ft = 96 in
The modular ratio of the slab and beam is
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n EE
ksiksi
cs
cg= = =4067
48880 832.
Therefore, the transformed flange width is
(bi)(n) = (96 in)(0.832) = 79.9 in
Transformed flange area is
(bin)(ts) = (79.9 in)(7.5 in) = 599.25 in (600.0 in2)
The area of transformed gross composite section, Agc, is
Agc = Ag + (bi)(n)(ts)
= 816 in2 + 600.0 in2
= 1416 in2
Use the transformed flange area to compute composite section properties by summing moments of areas about the centroid of the basic beam section.
0.5 in integralwearing thickness
96 in
8 intotal slabthickness
7.5 ineffective slab
thickness
14 in 18 in
17 in
10 in
9 in
c.g. basic beam
c.g. of compositesectiony =13.74 in
yb = 39.05 in
yt = 22.45 in
8 in 8 in
h = 54 iny t = 2
8.69
iny b =
25.
31 in
24 in
4 in5 in
0 in haunch used forcomposite section properties
–
FIGURE 3.32Composite section.
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A y in y in y ingc t( ) = ( ) = ( ) +
1416 600 7 52
2 2 .
= ( ) +( )600 28 69 3 752in in in. .
= 19 464 3, in
y = 13.74 in from the centroid of the basic beam.
Thus, for the composite section,
′ = + = + =y y y in in inbb b 25 31 13 74 39 05. . .
′ = +( ) − ′ = +( ) −y h t y in in intt s bb 54 0 7 5 39 05. . .
= 22.45 in
The composite moment of inertia about the centroid of the composite sec-tion is
I I A y b nt in y tc g g
i stt
s= + + + ( ) ′ −
23
22
12600
2
= + ( )( ) + ( )(255 194 816 13 7496 0 8324 2 2, .
.in in in
in ))( )
+ ( ) −
7 512
600 22 45 7 52
3
22
.
. .
in
in in in
= 621,867 in4
The section modulus of the composite section for the bottom extreme fiber is
S Iy
inin
incbc
bb=
′= =621 867
39 0515 925
43,
.,
The section modulus of the composite section for the top extreme fiber is
S Iy
inin
inctc
tt=
′= =621 867
22 4527 700
43,
.,
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Step 2: Perform the Dead Load Analysis
The beam weight is
w in ftin
kipsft
= ( )
=816 112
0 15 022
3. .. /85 kips ft
The dead weight of the slab is
w ft kipsft
in ftinD = ( )
( )
8 0 15 7 5 1123. .
= 0 75. /kips ft
wDC1 = w + wD
= 0.85 kips/ft + 0.75 kips/ft = 1.60 kips/ft
The superimposed dead load, ws, consists of the parapet and curb loads, distributed equally to the 6 beams.
wkips
ftbeams
s =
( )
=
0 506 2
60 169. . kkips ft/
ws = wDC2 = 0.169 kips/ft
wDC = 1.60 kips/ft + 0.169 kips/ft = 1.769 kips/ft
The future wearing surface dead load, wFWS, is assumed to be equally dis-tributed to each girder.
w ft kipsft
kips ftFWS = ( )
=8 0 03 0 240. . /
wFWS = wDW = 0.240 kips/ft
The maximum dead load moments and shears are
M wL w ftw f ipsmax = = ( ) =
2 2
8808
800 t-k
V wL w ftw f ipsmax = = ( ) =
2802
40 t-k
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Step 3: Calculate the Live Load Force Effects for Moment and Shear.
Where w is the clear roadway width between curbs and/or barriers. The number of lanes is
A Art. 3.6.1.1.1
N w ftL = = =
124812
4 (4 lanes)
A Tbl. 3.6.2.1-1
Please see Table 3.8.The distance between the centers of gravity of the basic beam and the deck is
e y t in in ing ts= + = + =2
28 69 7 52
32 44. . .
The area of the basic beam, A, is 816 in2.The moment of inertia of the basic beam, Ig, is 255,194 in4.The modular ratio between the beam and slab is
nEE
ksiksi
cg
cs= = =4888
40671 2.
The longitudinal stiffness parameter isA Eq. 4.6.2.2.1-1
Kg = n(Ig + Aeg2) = (1.2)(255,194 in4 + (816 in2)(32.44 in)2)
= 1,336,697.4
Distribution factor for moments per lane in interior beams with concrete deck,
A Art. 4.2.2.2b
TABLE 3.8
Dynamic Load Allowance, IM
Component IM (%)
Deck joints, all limit states 75Fatigue and fracture limit state 15All other limit states 33
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KLt ft ing
s121 336 697 4
12 80 7 53 303 3=
( )( )=, , .
..
Common deck type (j)
A Tbl. 4.6.2.2.1-1
The multiple presence factors apply to the lever rule case only. They have been included in the approximate equations for distribution factors in Arts. 4.6.2.2 and 4.6.2.3 for both single and multiple lanes loaded.
A Comm. 3.6.1.1.2
The distribution factor for moment for interior beams per lane with one design lane loaded is
A Tbl. 4.6.2.2.2b-1 or Appendix A
DFM S SL
KLtsig
s= +
0 06
14 12
0 4 0 3
3.. .
0 1.
= +
( )0 06 814
880
3 300 4 0 3
0. .. .
ft ftft
..1
= 0.51 lanes per beam
The distribution factor for moment for interior beams with two or more design lanes loaded is
DFM S SL
KLtmig
s= +
0 0759 5 12
0 6 0 2
..
. .
33
0 1
.
= +
(0 075 89 5
880
3 300 6 0 2
..
.. .
ft ftft ))0 1.
= 0.716 lanes per beam [controls]
The distribution factor for shear for interior beams (one design lane loaded) is
A Art. 4.6.2.2.3; Tbl. 4.6.2.2.3a-1 or Appendix C
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DFV S ftsi = +
= +
0 3625
0 36 8 025
. . .
= 0.68 lanes per beam
The distribution factor for shear for interior beams (two or more design lanes loaded) is
DFV S S ftmi = +
−
= + 0 212 35
0 2 8 012
2
. . .
−
835
2ft
= 0.814 lanes per beam [controls]
Step 4: Determine the Maximum Live Load Moments and Shears
Approximate the maximum bending moment at midspan due to the HL-93 loading. Please see Figures 3.33 through 3.36.
40 ft40 ft26 ft
A
14 ft4 ft
13 13
18 20
14 ft 26 ft
FIGURE 3.33Influence line diagram for maximum moment at midspan.
80 ft
Truck
32 kips
14 ft26 ft
A B
32 kips 8 kips
26 ft14 ft
FIGURE 3.34Design truck (HS-20) position for moment at midspan.
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Using the influence line diagram,
Mtr = (32 kips)(20 ft) + (32 kips + 8 kips)(13 ft)
= 1160 ft-kips [controls]
Mtandem = (25 kips)(20 ft + 18 ft) = 950 ft-kips
M
kipsft
ftipsln
.=
( )
=0 64 80
8512
2
ft-k
The maximum live load plus impact moment is defined by the following equation.
M DFM M or M IM MLL IM mi tr dem+ = ( ) +
+tan ln1100
= (0.716)((1160 ft-kips)(1 + 0.33) + 512 ft-kips)
= 1471.2 ft-kips
For the approximate maximum shear due to HL-93 loading, please see Figures 3.37 through 3.40, and use the influence line diagram.
Vtr = (32 kips)(1 + 0.825) + (8 kips)(0.65) = 63.60 kips [controls]
Vtandem = (25 kips)(1 + 0.95) = 48.75 kips
Tandem
25 kips 25 kips
40 ft 4 ft
A B
36 ft
FIGURE 3.35Design tandem load position for moment at midspan.
Lane
w = 0.64 kips/ft
A B
FIGURE 3.36Design lane load for moment at midspan.
274 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
V
kipsft
ftkipsln
..=
( )
=0 64 80
225 6
The maximum live load plus impact shear is defined by the following equation:
80 ft??14 ft14 ft
4 ft
1.0 0.950.825
0.65
FIGURE 3.37Influence line diagram for maximum shear at support.
Truck
14 ft14 ft
32 kips 32 kips 8 kips
a b
80 ft
FIGURE 3.38Design truck position for shear at support.
Tandem
25 kips25 kips4 ft
a b
FIGURE 3.39Design tandem load position for shear at support.
Lane
w = 0.64 kips/ft
a b
FIGURE 3.40Design lane load for shear at support.
275Practice Problems
© 2008 Taylor & Francis Group, LLC
V DFV V or V IM VLL IM tr dem+ = ( ) +
+ tan ln1
100
= (0.814)((63.6 kips)(1 + 0.33) + 25.6 kips) = 89.69 kips
Interior Beam Summary (using the maximum dead and live/impact load, shear, and moment equations)
Note that DC = DC1 + DC2 or wDC = wDC1 + wDC2
VDC = Vmax = 40 wDC = (40)(1.769 kips/ft) = 70.76 kips
MDC = Mmax = 800 wDC = (800)(1.769 kips/ft) = 1415.2 ft-kips
Due to the slab and the beam weight,
VDC1 = 40 wDC1 = (40)(1.60 kips/ft) = 64.0 kips
MDC1 = 800 wDC1 = (800)(1.60 kips/ft) = 1280 ft-kips
Due to superimposed dead loads – parapet and curb loads,
VDC2 = 40 wDC2 = (40)(0.169 kips/ft) = 6.76 kips
where:DC1 = beam weight + slab weight = 0.85 kips/ft + 0.75 kips/ft = 1.60 kips/ftDC2 = superimposed parapet and curb loads = 0.169 kips/ftMDC2 = 800 wDC2 = (800)(0.169 kips/ft) = 135.2 ft-kips
Due to the wearing surface weight,
VDW = 40 wDW = (40)(0.240 kips/ft) = 9.60 kips
MDW = 800 wDW = (800)(0.240 kips/ft) = 192.0 ft-kips
Please see Table 3.9.Recall that:
w = beam weight = 0.85 kips/ft
wD = slab weight = 0.75 kips/ft
wDC1 = w + wD = weights of beam and slab = 1.60 kips/ft
MDC1 = 1280.0 ft-kips
276 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
VDC1 = 64.0 kipsws = wDC2 = superimposed dead loads (parapet and curb) = 0.169 kips/ftMDC2 = 135.2 ft-kipsVDC2 = 6.76 kipswDC = wDC1 + wDC2 = weight of beam and slab plus parapet and curb =
1.769 kips/ftMDC = 1415.2 ft-kipsVDC = 70.76 kipswDW = wFWS = future wearing surface = 0.24 kips/ftMDW = 192.0 ft-kipsVDW = 9.6 kipsSncb = section modulus of noncomposite section for the bottom extreme
fiberSnct = section modulus of noncomposite section for the top extreme fiberScb = section modulus of composite section for the bottom extreme fiberSct = section modulus of composite section for the top extreme fiber
Step 5: Estimate the Required Prestress
Calculate the bottom tensile stress in prestressed concrete, fgb, noncomposite, and composite properties.
Service III Limit State governs for longitudinal analysis relating to tension in prestressed concrete superstructures with the objective of crack control and to principal tension in the webs of segmental concrete girders.
A Art. 3.4.1
Service III Limit StateA Tbl. 3.4.1-1
Q = η((1.0)(DC + DW) + (0.8)(LL + IM))
TABLE 3.9
Summary of Dead Load Moments and Shears
Load Type w (kips/ft) M (ft-kips) V (kips)
DC 1.769 1,415.2 70.76DC1 1.60 1,280.0 64.0DC2 0.169 135.2 6.76DW 0.240 192.0 9.60LL + IM N/A 1,471.2 89.69
277Practice Problems
© 2008 Taylor & Francis Group, LLC
η = 1.0
A Art. 1.3.2
f MS
M M MSgb
DC
ncb
DC DW LL IM
cb= + + + +1 2 0 8. ( )
=( )
+
1280 0 12
10 083
135 2
3
.
,
.
f ips inft
in
t-k
ff ips f ips f ipst-k t-k t-k+ +192 0 8 1471 215 92
. ( . ), 55 3in
= 2.66 (2.7 ksi)
Tensile stress limit at service limit state after losses, fully pretensioned is
A Art. 5.9.4.2.2; Tbl. 5.9.4.2.2-1
0 19 0 19 6 5 0 484. . . .′ = =f ksi ksicg
The excess tension in the bottom fiber due to applied loads is
f f f ksi ksi ksit gb cg= − ′ = − =0 19 2 66 0 484 2 18. . . .
The location of the center of gravity of the strands at midspan usually ranges from 5 to 15% of the beam depth. In this example, first assume 10%.
The distance between the center of gravity of the bottom strands to the bottom fiber is assumed to be
ybs = (0.1)(54 in) = 5.4 in
The strand eccentricity at midspan is
ec = yb – ybs = 25.31 in – 5.4 in = 19.91 in
To find the initial required prestress force, Pi,
278 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
f PA
PeSt
i
g
i c
ncb= +
2 18816
19 9110 0832 3.
.,
ksi Pin
P inin
i i= + ( )
Pi = 681.2 kips
The stress in prestressing tendon immediately prior to transfer, fpbt, is
A Art. 5.9.3
fpbt = 0.75 fpu = (0.75)(270 ksi) = 202.5 ksi
A Tbl. 5.9.3-1
Assuming a 25% prestress loss, prestress force per ½ in strand after losses, fps, is
fps = Apsfpbt(1 – 0.25) = (0.153 in2)(202.5 ksi)(1 – 0.25) = 23.2 kips
Number of strands required is
Pf
kipskips
i
ps= =681 2
23 229 4.
..
Try 30, ½ in strands.
Check the assumption of bottom strand center of gravity using the follow-ing configuration. Please see Figure 3.41.
The distance between the center of gravity of the 30 strands to the bottom fiber is
yin in in in
bs = ( )( ) + ( )( ) + ( )( ) + ( )( ) + ( )8 2 8 4 6 6 4 8 2 110 2 1230
in instrands
( ) + ( )( )
= 5.33 in
Inasmuch as the assumed value for ybs was 5.4 in, another iteration is not required.
The final strand eccentricity value is
ec = yb – ybs = 25.31 in – 5.33 in = 19.98 in
279Practice Problems
© 2008 Taylor & Francis Group, LLC
Revised initial required prestress force using 30 strands as opposed to 29.4 strands.
number of strands Pf
i
ps=
3023 2
strands Pkipsi=
.
Pi = 696 kips
Area of prestressing steel is
(30 strands)(0.153 in2) = 4.59 in2
Step 6: Calculate the Prestress Losses
In pretensioned members, ΔfpT = ΔfpES + ΔfpLT
A Eq. 5.9.5.1-1
ΔfpT total loss kips/in2
ΔfpES sum of all losses or gains due to elastic shortening or extension at the time of application of prestress and/or external loadsΔfpLT losses due to long-term shrinkage and creep of concrete, and relaxation of the steel kips/in2
96 in
dp = 61.5 in
7.5 in
c.g. basic beam
c.g. of 30strands
c.g. of 30strands
rows at 2 inspacing
g = ybs = 5.33 in
ec
y b =
25.
31 in
Slab
FIGURE 3.41Prestressed I-beam with 30, ½ in strands.
280 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Temporary allowable concrete stresses before losses due to creep and shrinkage are as follows (i.e., at time of initial prestress).Find the loss due to elastic shortening, ΔfpES.
A Eq. 5.9.5.2.3a-1
∆fEE
fpESp
cicgp=
fcgp is the concrete stress at the center of gravity of prestressing tendons due to prestressing force immediately after transfer and self-weight of mem-ber at the section of maximum moment.
fcgp is found using an iterative process. Alternatively, the following calcu-lation can be used to find ΔfpES for loss in prestressing steel due to elastic shortening.
A Eq. C5.9.5.2.3a-1
∆fA f I e A e M A
A I e ApES
ps pbt g m g m g g
ps g m g
=+( ) −
+( )2
2 ++A I E
Eg g ci
p
Aps area of prestressing steel 4.59 in2
Ag gross area of section 816 in2
Eci modulus of elasticity of concrete at transfer 4496 ksi
Ep modulus of elasticity of prestressing tendons 28,500 ksi
em = ec average prestressing steel eccentricity at midspan 19.98 in
Ig moment of inertia of the basic beam section 255,194 in4
The midspan moment due to beam self weight is
M wLkips
ftft
g = =
( )2
2
8
0 85 80
8
.
= 680 ft-kips
The loss due to elastic shortening is
A Eq. C5.9.5.2.3a-1
281Practice Problems
© 2008 Taylor & Francis Group, LLC
∆fA f I e A e M A
A I e ApES
ps pbt g m g m g g
ps g m g
=+( ) −
+( )2
2 ++A I E
Eg g ci
p
=
( )( ) + ( )4 59 202 5 255 194 19 98 812 4 2. . , .in ksi in in 66
19 98 680 121
2in
in f ips inft
( )( )−( )( )
. t-k ( )( ) + ( )
816
4 59 255 194 19 98 816
2
2 4 2
in
in in in. , . iin
in in ksi
2
2 4816 255 194 449628 50
( )( )+
( )( )( ),, 00 ksi
= 11.46 ksi
Calculate the long-term prestress loss due to creep of concrete shrinkage of concrete, and relaxation of steel, ΔfpLT.
A Art. 5.9.5.3; Eq. 5.9.5.3-1
Where:fpi = fpbt = stress in prestressing steel immediately prior to transferΔfpR = relaxation loss
ΔfpR is an estimate of relaxation loss taken as 2.4 ksi for low relaxation strand, 10.0 ksi for stress relieved strand, and in accordance with manufac-turer’s recommendation for other types of strand.
A Art. 5.9.5.3
γh = correction factor for relative humidity
γh = 1.7 – 0.01 H = 1.7 – (0.01)(0.72) = 1.69A Eq. 5.9.5.3-2
The average annual ambient relative humidity, H, in western Pennsylvania is approximately 72%.
A Fig. 5.4.2.3.3-1
The correction factor for specified concrete strength at the time of the pre-stress transfer to the concrete is,
A Eq. 5.9.5.3-3
282 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
γ stcif
=+ ′( ) =
+( )5
15
1 5 5.
= 0.77
∆ ∆ff A
AfpLT
pi ps
gh st h st pR= ( ) + ( ) +10 0 12 0. .γ γ γ γ
A Eq. 5.9.5.3-1
= ( ) ( )( )
(10 0
202 5 4 59816
1 692
2.. .
.ksi in
in ))( ) + ( )( )( ) +0 77 12 1 69 0 77 2 4. . . . ksi
= 32.84 ksi
The total prestress losses are
ΔfpT = ΔfpES + ΔfpLT = 11.46 ksi + 32.84 ksi = 44.30 ksi
A Eq. 5.9.5.1-1
Summary of Prestress Forces
The prestress stress per strand before transfer fpbt is
A Tbl. 5.9.3-1
fpbt = 0.75 fpu
= ( )
=0 75 270 202 52 2. .kipsin
kipsin
The prestress force per strand before transfer is
Ppi = (fpbt)Aps
= (202.5 ksi)(0.153 in2) = 31 kips
The prestress force per strand after all losses is
Ppe = (202.5 ksi – 44.30 ksi)(0.153 in2) = 24.20 kips
Check assumption of 25% prestress losses.
283Practice Problems
© 2008 Taylor & Francis Group, LLC
% .lossprestress force after all losses Ppe= −
=(1 0 ))=( ) = −
prestress force before transfer Ppi1 24 2. 00
3121 9kips
kips= . %
Assumption of 25% was conservative because 21.9% < 25% [OK].
Step 7: Check Tensile and Compressive Concrete Stresses at Transfer
The compressive stress of concrete at time of prestressing before losses is
A Art. 5.9.4.1.1
fci = 0.60 f′ci = (0.6)(5.5 ksi) = 3.3 ksi
The tensile stress in prestressed concrete before losses is
A Art. 5.9.4.1.2; Tbl. 5.9.4.1.2-1
f f ksi ksiti ci= ′ = =0 24 0 24 5 5 0 563. . . .
Stress at bottom of girder (compressive stress)
f PA
PeS
MSbot
i
g
i c
ncb
g
ncb= − − +
= − − ( )( )696816
696 19 9810 0832 3
kipsin
kips inin
.,
++
68010 083
123
f psin
inft
t-ki,
= –1.42 ksi < –3.3 ksi [OK]
Stress at top of girder (tensile stress)
f PA
PeS
MSbot
i
g
i c
nct
g
nct= − + −
= − + ( )( )696816
696 19 988895 02 3
kipsin
kips inin
..
−−
6808895 0
123
f psin
inft
t-ki.
= –0.207 ksi < –0.563 ksi [OK]
284 Simplified LRFD Bridge Design
© 2008 Taylor & Francis Group, LLC
Step 8: Determine the Flexural Resistance Using the Strength I Limit State
Strength I Limit State, U = 1.25 DC + 1.50 DW + 1.75(LL + IM).A Tbl. 3.4.1-1
Recall that DC is the weight due to the deck slab, the basic beam, and the parapet curb, and that DW is the weight due to the future wearing surface.
For these values, the Strength I Limit State can be written as follows:
Mu = (1.25)(1415.2 ft-kips) + (1.50)(192 ft-kips) + (1.75)(1471.2 ft-kips)
= 4631.6 ft-kips
Find the average stress in prestressing steel assuming rectangular behavior.A Eq. 5.7.3.1.1-1
f f k cdps pu
p= −
1
where:dp = distance from extreme compression fiber to the centroid of the pre-
stressing tendonsfpu = 270 ksik = 0.38
A Tbl. C5.7.3.1.1-1
dp = (h – ybs (= g) + ts) = (54 in – 5.33 in) + 7.5 in = 56.17 in
cA f A f A f
f b k Af
ps pu s s s s
c psp
=( )( ) + − ′ ′
′ + ( )0 85 1. β uu
pd
A Eq. 5.7.3.1.1-4
where: β1 0 85 0 05 4= − ′ −( ). . f ksicg
A Art. 5.7.2.2
= 0.85 – 0.05(6.5 ksi – 4 ksi)
= 0.725
285Practice Problems
© 2008 Taylor & Francis Group, LLC
c
in ksi
ksi in=
( )( )( )( )(
4 59 270
0 85 6 5 0 725 96
2.
. . . )) + ( )( )
0 38 4 59 27056 17
2. ..
in ksiin
= 3.16 in < ts = 7.5 in
so the assumption is OK.The average stress in prestressing steel when the nominal resistance of
member is required, fps,
f ksi ininps = ( ) − ( )
=270 1 0 38 3 1656 17
264. ..
..2 ksi
Find the factored flexural resistance for flanged section:A Art. 5.7.3.2.2
a = β1c = (0.725)(3.16 in)
= 2.29 in
The factored resistance Mr shall be taken as:A Eq. 5.7.3.2.2
Mr = ΦMn
where:Mn = nominal resistanceΦ = resistance factor = 1.0
A Art. 5.5.4.2
Mr = (1.0)(Mn) = Mn
A Eq. 5.7.3.2.2-1
M A f d an ps ps p= −
2
Mr = Mn
M in ksi in inr = ( )( ) −
4 59 264 2 56 17 2 292
2. . . . 1112
ftin
Mr = 5560 ft-kips > Mu = 4631.6 ft-kips [OK]
287© 2008 Taylor & Francis Group, LLC
Appendix A: Distribution of Live Loads per Lane for Moment in Interior Beams(AASHTO Table 4.6.2.2.2b-1)
288 Appendix A
© 2008 Taylor & Francis Group, LLC
Dis
trib
utio
n of
Liv
e L
oad
s p
er L
ane
for
Mom
ent i
n In
teri
or B
eam
s
Typ
e of
Su
per
stru
ctu
reA
pp
lica
ble
Cro
ss S
ecti
on
from
Tab
le 4
.6.2
.2.1
-1D
istr
ibu
tion
Fac
tors
Ran
ge o
f A
pp
lica
bil
ity
Woo
d D
eck
on W
ood
or
Ste
el B
eam
sa,
1Se
e T
able
4.6
.2.2
.2a-
1
Con
cret
e D
eck
on
Woo
d B
eam
s1
One
Des
ign
Lan
e L
oad
ed:
S/12
.0Tw
o or
Mor
e D
esig
n L
anes
Loa
ded
:S/
10.0
S ≤
6.0
Con
cret
e D
eck,
Fill
ed
Gri
d, P
artia
lly F
illed
G
rid,
or
Unfi
lled
Gri
d
Dec
k C
ompo
site
with
R
einf
orce
d C
oncr
ete
Slab
on
Stee
l or
Con
cret
e Be
ams;
C
oncr
ete
T-Be
ams,
T-
and
Dou
ble
T-Se
ctio
ns
a, e
, k, a
nd a
lso
i, j i
f su
ffici
ently
con
nect
ed to
ac
t as
a un
it
One
Des
ign
Lane
Loa
ded:
006
1412
0
04
03
3.
.
..
+
S
S LK
Ltg
s0
1.
Two
or M
ore
Des
ign
Lan
es L
oad
ed:
007
59
512
0
06
02
3.
..
..
+
SS L
KLtg
s
01.
3.5
≤ S
≤ 16
.0
4.5
≤ t s
≤ 12
.0
20 ≤
L ≤
240
Nb ≥
4
10,0
00 ≤
Kg ≤
7,0
00,0
00
Use
less
er o
f the
val
ues
obta
ined
from
the
equa
tion
abov
e w
ith N
b = 3
or
the
leve
r ru
leN
b = 3
Cas
t-in
-Pla
ce C
oncr
ete
Mul
tice
ll B
oxd
One
Des
ign
Lan
e L
oad
ed:
175
36
11
035
04 5
..
..
+
SL
Nc
Two
or M
ore
Des
ign
Lan
es L
oad
ed:
135
81
03
025
NS
Lc
..
.
7.0
≤ S
≤ 13
.0
60 ≤
L ≤
240
Nc ≥
3
If N
c > 8
use
Nc =
8
289Appendix A
© 2008 Taylor & Francis Group, LLC
Con
cret
e D
eck
on
Con
cret
e Sp
read
Box
B
eam
s
b, c
One
Des
ign
Lan
e L
oad
ed:
SSd
L3
012
0
035
2
025
..
..
Two
or M
ore
Des
ign
Lan
es L
oad
ed:
SSd
L6
312
0
06
2
012
5
..
..
6.0
≤ S
≤ 18
.0
20 ≤
L ≤
140
18 ≤
d ≤
65
Nb ≥
3
Use
Lev
er R
ule
S >
18.
0
Con
cret
e B
eam
s us
ed
in M
ulti
beam
Dec
ksf
One
Des
ign
Lan
e L
oad
ed:
kb
LI J
333
05
025
.
..
whe
re: k
= 2
.5(N
b)−
0.2 ≥
1.5
Two
or M
ore
Des
ign
Lan
es L
oad
ed:
kb
bL
I J30
512
0
06
02
006
..
.
.
35 ≤
b ≤
60
20 ≤
L ≤
120
5 ≤
Nb ≤
20
g if
suf
fici
entl
y co
nnec
ted
to
act
as
a un
it
Sour
ce:
AA
SH
TO
Tab
le 4
.6.2
.2.2
b-1
.
291© 2008 Taylor & Francis Group, LLC
Appendix B: Distribution of Live Loads per Lane for Moment in Exterior Longitudinal Beams(AASHTO Table 4.6.2.2.2d-1)
292 Appendix B
© 2008 Taylor & Francis Group, LLC
Dis
trib
utio
n of
Liv
e L
oad
s p
er L
ane
for
Mom
ent i
n E
xter
ior
Lon
gitu
din
al B
eam
s
Typ
e of
Su
per
stru
ctu
reA
pp
lica
ble
Cro
ss S
ecti
on
from
Tab
le 4
.6.2
.2.1
-1
On
e D
esig
n L
ane
Loa
ded
Two
or M
ore
Des
ign
L
anes
Loa
ded
Ran
ge o
f A
pp
lica
bil
ity
Woo
d D
eck
on W
ood
or
Stee
l B
eam
sa,
1L
ever
Rul
eL
ever
Rul
eN
/A
Con
cret
e D
eck
on W
ood
B
eam
s1
Lev
er R
ule
Lev
er R
ule
N/
A
Con
cret
e D
eck,
Fill
ed G
rid
, Pa
rtia
lly F
illed
Gri
d, o
r U
nfille
d G
rid
Dec
k C
ompo
site
w
ith
Rei
nfor
ced
Con
cret
e Sl
ab
on S
teel
or
Con
cret
e B
eam
s;
Con
cret
e T-
Bea
ms,
T- a
nd
Dou
ble
T-Se
ctio
ns
a, e
, k, a
nd a
lso
i, j i
f su
ffici
entl
y co
nnec
ted
to a
ct
as a
uni
t
Lev
er R
ule
g =
e g i
nter
ior
ed e
=+
077
91
..
−1.0
≤ d
e ≤ 5
.5
Use
less
er o
f the
val
ues
obta
ined
fr
om th
e eq
uati
on a
bove
wit
h N
b = 3
or
the
leve
r ru
le
Nb =
3
Cas
t-in
-Pla
ce C
oncr
ete
Mul
tice
ll B
oxd
gW
e=
14g
We
=14
We ≤
S
or th
e pr
ovis
ions
for
a w
hole
-wid
th d
esig
n sp
ecifi
ed in
Art
icle
4.6
.2.2
.1
Con
cret
e D
eck
on C
oncr
ete
Spre
ad B
ox B
eam
sb,
cL
ever
Rul
eg
= e
gin
teri
or
ed e
=+
097
285
..
0 ≤
de ≤
4.5
6.0
< S
≤ 1
8.0
Use
Lev
er R
ule
S >
18.
0
293Appendix B
© 2008 Taylor & Francis Group, LLC
Con
cret
e B
ox B
eam
s U
sed
in
Mul
tibe
am D
ecks
f, g
g =
e g
inte
rior
ed e
=+
≥1
125
301
0.
.
g =
e g
inte
rior
ed e
=+
≥1
0425
10
..
de ≤
2.0
Con
cret
e Be
ams
Oth
er th
an B
ox
Beam
s U
sed
in M
ultib
eam
D
ecks
hLe
ver R
ule
Leve
r Rul
eN
/A
i, j i
f con
nect
ed o
nly
enou
gh
to p
reve
nt r
elat
ive
vert
ical
d
ispl
acem
ent a
t the
inte
rfac
e
Ope
n St
eel G
rid
Dec
k on
Ste
el
Bea
ms
aL
ever
Rul
eL
ever
Rul
eN
/A
Con
cret
e D
eck
on M
ulti
ple
Stee
l Box
Gir
der
sb,
cA
s sp
ecifi
ed in
Tab
le 4
.6.2
.2.2
b-1
Sour
ce:
AA
SH
TO
Tab
le 4
.6.2
.2.2
d-1
.
295© 2008 Taylor & Francis Group, LLC
Appendix C: Distribution of Live Load per Lane for Shear in Interior Beams(AASHTO Table 4.6.2.2.3a-1)
296 Appendix C
© 2008 Taylor & Francis Group, LLC
Dis
trib
utio
n of
Liv
e L
oad
per
Lan
e fo
r Sh
ear
in I
nter
ior
Bea
ms
Typ
e of
Su
per
stru
ctu
reA
pp
lica
ble
Cro
ss S
ecti
on
from
Tab
le 4
.6.2
.2.1
-1
On
e D
esig
n L
ane
Loa
ded
Two
or M
ore
Des
ign
Lan
es
Loa
ded
Ran
ge o
f A
pp
lica
bil
ity
Woo
d D
eck
on W
ood
or
Stee
l Bea
ms
a, 1
See
Tab
le 4
.6.2
.2.2
a-1
Con
cret
e D
eck
on W
ood
Bea
ms
1L
ever
Rul
eL
ever
Rul
eN
/A
Con
cret
e D
eck,
Fill
ed G
rid
, Par
tial
ly
Fille
d G
rid
, or
Unfi
lled
Gri
d D
eck
Com
posi
te w
ith
Rei
nfor
ced
Con
cret
e Sl
ab o
n St
eel o
r C
oncr
ete
Bea
ms;
C
oncr
ete
T-B
eam
s, T
- and
Dou
ble
T-Se
ctio
ns
a, e
, k, a
nd a
lso
i, j i
f su
ffici
entl
y co
nnec
ted
to
act a
s a
unit
036
250
..
+S
02
1235
20
..
+
S
S3.
5 ≤
S ≤
16.0
20 ≤
L ≤
240
4.5
≤ t s
≤ 12
.0N
b ≥ 4
Lev
er R
ule
Lev
er R
ule
Nb =
3
Cas
t-in
-Pla
ce C
oncr
ete
Mul
tice
ll B
oxd
Sd
L9
512
0
06
01
..
..
Sd
L7
312
0
09
01
..
..
6.0
≤ S
≤ 13
.020
≤ L
≤ 2
4035
≤ d
≤ 1
10N
c ≥ 3
Con
cret
e D
eck
on C
oncr
ete
Spre
ad
Box
Bea
ms
b, c
Sd
L10
120
06
01
..
.S
dL
74
120
08
01
..
..
6.0
≤ S
≤ 18
.020
≤ L
≤ 1
4018
≤ d
≤ 6
5N
b ≥ 3
Lev
er R
ule
Lev
er R
ule
S >
18.
0
297Appendix C
© 2008 Taylor & Francis Group, LLC
Con
cret
e B
ox B
eam
s U
sed
in
Mul
tibe
am D
ecks
f, g
bL
I J13
0
015
005
..
bb
LI J
b15
612
048
04
01
005
..
.
.
≥b 48
10.
35 ≤
b ≤
60
20 ≤
L ≤
120
5 ≤
Nb ≤
20
25,0
00 ≤
J ≤
610,
000
40,0
00 ≤
I ≤
610,
000
Con
cret
e B
eam
s O
ther
Tha
n B
ox
Bea
ms
Use
d in
Mul
tibe
am D
ecks
h i, j i
f con
nect
ed o
nly
enou
gh to
pre
vent
rel
ativ
e ve
rtic
al d
ispl
acem
ent a
t th
e in
terf
ace
Lev
er R
ule
Lev
er R
ule
N/
A
Ope
n St
eel G
rid
Dec
k on
Ste
el
Bea
ms
aL
ever
Rul
eL
ever
Rul
eN
/A
Con
cret
e D
eck
on M
ulti
ple
Stee
l Box
B
eam
sb,
cA
s sp
ecifi
ed in
Tab
le 4
.6.2
.2.2
b-1
Sour
ce:
AA
SH
TO
Tab
le 4
.6.2
.2.3
a-1.
299© 2008 Taylor & Francis Group, LLC
Appendix D: Distribution of Live Load per Lane for Shear in Exterior Beams(AASHTO Table 4.6.2.2.3b-1)
300 Appendix D
© 2008 Taylor & Francis Group, LLC
Dis
trib
utio
n of
Liv
e L
oad
per
Lan
e fo
r Sh
ear
in E
xter
ior
Bea
ms
Typ
e of
Su
per
stru
ctu
reA
pp
lica
ble
Cro
ss S
ecti
on
from
Tab
le 4
.6.2
.2.1
-1
On
e D
esig
n L
ane
Loa
ded
Two
or M
ore
Des
ign
Lan
es
Loa
ded
Ran
ge o
f A
pp
lica
bil
ity
Woo
d D
eck
on W
ood
or
Stee
l Bea
ms
a, 1
Lev
er R
ule
Lev
er R
ule
N/
A
Con
cret
e D
eck
on W
ood
Beam
s1
Leve
r R
ule
Leve
r R
ule
N/A
Con
cret
e D
eck,
Fill
ed G
rid
, Par
tial
ly
Fille
d G
rid
, or
Unfi
lled
Gri
d D
eck
Com
posi
te w
ith
Rei
nfor
ced
Con
cret
e Sl
ab
on S
teel
or
Con
cret
e B
eam
s; C
oncr
ete
T-B
eam
s, T
- and
Dou
ble
T-Se
ctio
ns
a, e
, k, a
nd a
lso
i, j i
f su
ffici
entl
y co
nnec
ted
to a
ct
as a
uni
t
Lev
er R
ule
g =
e g
inte
rior
ed e
=+
06
10.
–1.0
≤ d
e ≤ 5
.5
Lev
er R
ule
Nb =
3
Cas
t-in
-Pla
ce C
oncr
ete
Mul
tice
ll B
oxd
Lev
er R
ule
g =
e g
inte
rior
ed e
=+
064
125
..
–2.0
≤ d
e ≤ 5
.0
or th
e pr
ovis
ions
fo
r a
who
le-w
idth
d
esig
n sp
ecifi
ed
in A
rtic
le 4
.6.2
.2.1
Con
cret
e D
eck
on C
oncr
ete
Spre
ad B
ox
Bea
ms
b, c
Lev
er R
ule
g =
e g
inte
rior
ed e
=+
08
10.
0 ≤
de ≤
4.5
Lev
er R
ule
S >
18.
0
301Appendix D
© 2008 Taylor & Francis Group, LLC
Con
cret
e B
ox B
eam
s U
sed
in M
ulti
beam
D
ecks
f, g
g =
e g
inte
rior
ed e
=+
125
201
0.
.≥
geg
b
b ed
be
=
≤
=+
+−
inte
rior
48
481
0
112
20
4
.
.
00
05
.
≥1.
0
de ≤
2.0
35≤
b ≤
60
Con
cret
e B
eam
s O
ther
than
Box
Bea
ms
Use
d in
Mul
tibe
am D
ecks
hL
ever
Rul
eL
ever
Rul
eN
/A
i, j i
f con
nect
ed o
nly
enou
gh
to p
reve
nt r
elat
ive
vert
ical
d
ispl
acem
ent a
t the
inte
rfac
e
Ope
n St
eel G
rid
Dec
k on
Ste
el B
eam
sa
Lev
er R
ule
Lev
er R
ule
N/
A
Con
cret
e D
eck
on M
ulti
ple
Stee
l Box
B
eam
sb,
cA
s sp
ecifi
ed in
Tab
le 4
.6.2
.2.2
b-1
Sour
ce:
AA
SH
TO
Tab
le 4
.6.2
.2.3
b-1
.
303© 2008 Taylor & Francis Group, LLC
Appendix E: U.S. Customary Units and Their SI Equivalents
U.S. Customary Units and Their SI Equivalents
Quantity U.S. Customary Units SI Equivalent
Area ft2 0.0929 m2
in2 645.2 mm2
Force kip 4.448 kN
lbf 4.448 N
Length ft 0.3048 m
in 25.40 mm
Moment ft-lbf 1.356 N-m
ft-kip 1.355 × 10 4 kN-m
in-lbf 0.1130 N-m
in-kip 1.130 × 104 kN-m
Moment of Inertia in4 0.4162 × 106 mm4
Stress lbf/ft2 47.88 Pa
lbf/in2 (psi) 6.895 kPa
kip/in2 (ksi) 6895.0 kPa
305© 2008 Taylor & Francis Group, LLC
References
Primary References
The following sources were used to create this book and can serve as pri-mary references for working bridge design-related problems for the NCEES civil and structural PE exams. (If a reference is referred to in this book by a name other than its title, the alternative name is indicated in brackets follow-ing the reference.)American Association of State Highway and Transportation Officials. AASHTO LRFD
Bridge Design Specifications, 5th ed., Washington, DC2010. [AASHTO]American Association of State Highway and Transportation Officials. Manual for
Bridge Evaluation. 2nd ed. Washington, DC: 2011. [MBE-2]American Association of State Highway and Transportation Officials. AASHTO LRFD
Bridge Design Specifications, 4th ed. Washington, DC: 2007 (with 2008 Interim Revisions). [AASHTO]
American Association of State Highway and Transportation Officials. Standard Specifications for Highway Bridges. 17th ed. Washington, DC: 2002.
American Institute of Steel Construction. Steel Construction Manual. 13th ed. Chicago: 2005. [AISC Manual]
Supplementary References
The following additional references were used to create this book.Barker, Richard M., and Jay A. Pucket. Design of Highway Bridges, an LRFD Approach,
2nd edition, Hoboken: John Wiley & Sons, Inc. 2007.Kim, Robert H., and Jai B. Kim. Bridge Design for the Civil and Structural Professional
Engineering Exams, 2nd ed., Belmont, CA: Professional Publications, Inc. 2001.Kim, Robert H., and Jai B. Kim. Relocation of Coudersport 7th Street Bridge (his-
toric truss) to 4th Street over Allegheny River, Potter County, Pennsylvania. Unpublished report for Pennsylvania State Department of Transportation. 2007–2009.
Schneider, E.F., and Shrinivas B. Bhide. Design of concrete bridges by the AASHTO LRFD specifications and LRFD design of cast-in-place concrete bridges. Report for the Portland Cement Association. Skokie, IL: July 31–August 3, 2007.
Sivakumar, B. Design and evaluation of highway bridge superstructures using LRFD. Unpublished American Society of Civil Engineers seminar notes. Phoenix, AZ: 2008.
6000 Broken Sound Parkway, NW Suite 300, Boca Raton, FL 33487711 Third Avenue New York, NY 100172 Park Square, Milton Park Abingdon, Oxon OX14 4RN, UK
an informa business
www.taylorandfrancisgroup.com w w w . c r c p r e s s . c o m
K16288
CIVIL ENGINEERING
Jai B. KimRobert H. Kim
Jonathan R. Eberle
SIMPLIFIED LRFD BRIDGE DESIGN
SIMPLIFIED
LRFD BRID
GE D
ESIGN
Kim • Kim
• Eberle
With Eric J. Weaver and Dave M. Mante
Simplified LRFD Bridge Design is a study guide for solving bridge problems on the Civil and Structural PE exams. It is also suitable as a reference for practicing engineers and as a classroom text for civil engineering students. The book conforms to the fifth edition of AASHTO LRFD Bridge Design Specifications (2010).
Unlike most engineering books, Simplified LRFD Bridge Design uses an alternative approach to learning––the inductive method. The book introduces topics by presenting specific design examples, literally teaching backward––the theory is presented once specific design examples are comprehended.
Another unique quality of the book is that whenever new topics and materials appear in design examples, AASHTO LRFD Bridge Design Specifications reference numbers are cited, so that students will know where to find those new topics and materials.
For example,
New Topics or Material AASHTO Reference Number Cited
Design Live Load HL-93 A Art. 3.6.1.2
Design Examples and Practice ProblemsIn addition to the first section on an overview of the LRFD Method of Bridge Design, there are eight design examples and three practice problems utilizing a step-by-step process to help students learn easily in the shortest time.
About the EditorsJai B. Kim, PhD, PE, is a professor emeritus of civil and environmental engineering at Bucknell University, and was department chairman for 26 years. Recently he was a structural engineer at FHWA. He was also actively involved in the NCEES structural PE Committee and Transportation Research Board Committee of Bridges and Structures. He holds a BSCE and MSCE from Oregon State University and a PhD from University of Maryland. Robert H. Kim, MSCE, PE, is chief design engineer for BKLB Structural Consultants, Inc. He has extensive experience in bridge engineering. He holds a BS from Carnegie Mellon University and a MSCE from The Pennsylvania State University. Jonathan R. Eberle, BSCE and EIT, is engaged in research with a focus on the seismic design of structures at Virginia Polytechnic Institute. He holds a BSCE from Bucknell University.
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